401. A \(5\,\text{kg}\) block rests on a rough horizontal surface. A \(25\,\text{N}\) force acts at \(37^\circ\) above the horizontal towards the right, while another horizontal force of \(8\,\text{N}\) acts towards the left. In a second arrangement, the \(25\,\text{N}\) force is reflected below the horizontal while retaining the same horizontal direction and magnitude. Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). If \(\mu_1\) and \(\mu_2\) are the minimum coefficients of static friction required for equilibrium in the first and second arrangements, respectively, what is \(\mu_1/\mu_2\)?
ⓐ. \(\dfrac{7}{13}\)
ⓑ. \(\dfrac{13}{7}\)
ⓒ. \(\dfrac{35}{12}\)
ⓓ. \(\dfrac{65}{12}\)
Correct Answer: \(\dfrac{13}{7}\)
Explanation: The horizontal component of the \(25\,\text{N}\) force is
\[
25\cos37^\circ=20\,\text{N}.
\]
After including the \(8\,\text{N}\) leftward force, the net horizontal tendency is
\[
20-8=12\,\text{N}
\]
to the right in both arrangements. Therefore \(12\,\text{N}\) of leftward static friction is required at the limiting minimum-coefficient condition. In the first arrangement, the upward component is
\[
25\sin37^\circ=15\,\text{N},
\]
so
\[
N_1=50-15=35\,\text{N}.
\]
Hence,
\[
\mu_1=\frac{12}{35}.
\]
In the second arrangement the \(15\,\text{N}\) component acts downward:
\[
N_2=50+15=65\,\text{N}.
\]
Thus,
\[
\mu_2=\frac{12}{65}.
\]
Therefore,
\[
\frac{\mu_1}{\mu_2}
=\frac{12/35}{12/65}
=\frac{65}{35}
=\frac{13}{7}.
\]
402. A \(5\,\text{kg}\) block rests on a rough horizontal surface with \(\mu_s=0.50\). A constant \(8\,\text{N}\) horizontal force acts towards the left. Another force \(P\) acts \(37^\circ\) above the horizontal towards the right. Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). For what range of \(P\) can the block remain at rest, and when does the direction of static friction reverse?
ⓐ. \(0\le P\le30\,\text{N}\); friction reverses at \(P=10\,\text{N}\)
ⓑ. \(8\le P\le25\,\text{N}\); friction reverses at \(P=8\,\text{N}\)
ⓒ. \(0\le P\le25\,\text{N}\); friction is always leftward
ⓓ. \(10\le P\le30\,\text{N}\); friction is always rightward
Correct Answer: \(0\le P\le30\,\text{N}\); friction reverses at \(P=10\,\text{N}\)
Explanation: The upward component of \(P\) reduces the normal reaction:
\[
N=50-0.6P.
\]
The rightward component of \(P\) is
\[
0.8P.
\]
Ignoring friction, the net horizontal force is
\[
0.8P-8.
\]
Hence the static friction required for equilibrium has magnitude
\[
|0.8P-8|.
\]
The maximum available static friction is
\[
f_{s,\max}=0.50(50-0.6P)=25-0.3P.
\]
Therefore,
\[
|0.8P-8|\le25-0.3P.
\]
For \(P\ge0\), the restrictive condition gives
\[
0.8P-8\le25-0.3P,
\]
so
\[
1.1P\le33,
\qquad
P\le30\,\text{N}.
\]
The non-frictional horizontal forces balance when \(0.8P=8\), i.e. at \(P=10\,\text{N}\). Below this value friction acts rightward; above it friction acts leftward.
403. A \(5\,\text{kg}\) block is placed on a rough plane inclined at \(37^\circ\) above the horizontal. The coefficient of static friction is \(0.50\). A horizontal force \(P\) acts towards the rising side of the plane. Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). For what range of \(P\) can the block remain at rest?
ⓐ. \(20\,\text{N}\le P\le50\,\text{N}\)
ⓑ. \(0\le P\le37.5\,\text{N}\)
ⓒ. \(\dfrac{100}{11}\,\text{N}\le P\le100\,\text{N}\)
ⓓ. \(\dfrac{100}{11}\,\text{N}\le P\le37.5\,\text{N}\)
Correct Answer: \(\dfrac{100}{11}\,\text{N}\le P\le100\,\text{N}\)
Explanation: The component of weight down the plane is
\[
mg\sin37^\circ=50(0.6)=30\,\text{N}.
\]
The horizontal force contributes
\[
P\cos37^\circ=0.8P
\]
up the plane. It also presses the block into the plane, so
\[
N=50(0.8)+P(0.6)=40+0.6P.
\]
The static friction required for equilibrium is therefore
\[
f=30-0.8P,
\]
where the sign determines its direction. Static equilibrium requires
\[
|30-0.8P|\le0.50(40+0.6P).
\]
Thus,
\[
|30-0.8P|\le20+0.3P.
\]
The lower threshold follows from
\[
30-0.8P\le20+0.3P,
\]
giving
\[
P\ge\frac{100}{11}\,\text{N}.
\]
The upper threshold follows from
\[
-30+0.8P\le20+0.3P,
\]
giving
\[
P\le100\,\text{N}.
\]
Friction acts up the plane at smaller allowed \(P\), becomes zero at \(P=37.5\,\text{N}\), and acts down the plane for larger allowed \(P\).
404. A \(2\,\text{kg}\) block rests without slipping on the horizontal floor of a truck. The coefficient of static friction between the block and truck is \(0.40\). In addition, a constant horizontal force of \(4\,\text{N}\) acts on the block towards the right. The truck accelerates towards the right with acceleration \(a\). Take \(g=10\,\text{m s}^{-2}\). For what range of \(a\) can the block remain at rest relative to the truck?
ⓐ. \(0\le a\le8\,\text{m s}^{-2}\), with friction reversing at \(a=4\,\text{m s}^{-2}\)
ⓑ. \(0\le a\le4\,\text{m s}^{-2}\), with friction always rightward
ⓒ. \(2\le a\le6\,\text{m s}^{-2}\), with friction always leftward
ⓓ. \(0\le a\le6\,\text{m s}^{-2}\), with friction reversing direction at \(a=2\,\text{m s}^{-2}\)
Correct Answer: \(0\le a\le6\,\text{m s}^{-2}\), with friction reversing direction at \(a=2\,\text{m s}^{-2}\)
Explanation: Work in the ground frame and let rightward be positive. If the block does not slip, it must have the truck's acceleration \(a\). Let \(f\) be the signed friction on the block. Newton's Second Law gives
\[
4+f=2a.
\]
Therefore,
\[
f=2a-4.
\]
The largest available static-friction magnitude is
\[
f_{s,\max}=0.40(2)(10)=8\,\text{N}.
\]
Thus,
\[
|2a-4|\le8.
\]
For the stated rightward truck acceleration \(a\ge0\), this gives
\[
0\le a\le6\,\text{m s}^{-2}.
\]
At \(a\lt 2\,\text{m s}^{-2}\), \(f\lt 0\), so friction acts leftward. At
\[
a=2\,\text{m s}^{-2},
\]
friction is zero. For larger \(a\), friction acts rightward to provide the additional acceleration required by the block.
405. A \(2\,\text{kg}\) block \(A\) rests on a \(3\,\text{kg}\) block \(B\), and the floor below \(B\) is smooth. In Trial I, a \(15\,\text{N}\) force acts rightward on \(A\) while a \(5\,\text{N}\) force acts leftward on \(B\). In Trial II, the \(15\,\text{N}\) rightward force acts on \(B\) and the \(5\,\text{N}\) leftward force acts on \(A\). What minimum coefficient of static friction between \(A\) and \(B\) is required for no slipping in each trial?
ⓐ. Trial I: \(0.60\); Trial II: \(0.30\)
ⓑ. Trial I: \(0.45\); Trial II: \(0.55\)
ⓒ. Trial I: \(0.50\); Trial II: \(0.50\)
ⓓ. Trial I: \(0.55\); Trial II: \(0.45\)
Correct Answer: Trial I: \(0.55\); Trial II: \(0.45\)
Explanation: In both trials the net external horizontal force on the two-block system is
\[
15-5=10\,\text{N}.
\]
The total mass is \(5\,\text{kg}\), so if the blocks move together,
\[
a=\frac{10}{5}=2\,\text{m s}^{-2}.
\]
In Trial I, for block \(A\),
\[
15+f=2(2).
\]
Thus,
\[
f=-11\,\text{N},
\]
meaning \(11\,\text{N}\) leftward friction is required. The interface normal is
\[
N=2(10)=20\,\text{N},
\]
so
\[
\mu_{s,\min}=\frac{11}{20}=0.55.
\]
In Trial II,
\[
-5+f=2(2),
\]
so
\[
f=9\,\text{N}
\]
rightward. Hence,
\[
\mu_{s,\min}=\frac9{20}=0.45.
\]
The same system acceleration can therefore require different friction magnitudes and even opposite friction directions depending on where the external forces are applied.
406. The same pair of rough surfaces is tested under three normal loads. In Trial P, a tangential force of \(16\,\text{N}\) places the block just at impending motion when \(N=40\,\text{N}\). The observations in Trials Q and R are then recorded.
| Trial | Normal reaction | Applied tangential force | Recorded state |
|---|
| P | \(40\,\text{N}\) | \(16\,\text{N}\) | Just impending |
| Q | \(60\,\text{N}\) | \(20\,\text{N}\) | At rest |
| R | \(30\,\text{N}\) | \(14\,\text{N}\) | At rest |
Assuming the elementary static-friction model, which conclusion is correct?
ⓐ. \(\mu_s=0.40\); Q can remain at rest with \(f=20\,\text{N}\); R cannot remain at rest.
ⓑ. \(\mu_s=0.40\); Q is at limiting equilibrium with \(f=24\,\text{N}\); R is also at limiting equilibrium.
ⓒ. \(\mu_s=0.50\); Q can remain at rest with \(f=20\,\text{N}\); R cannot remain at rest.
ⓓ. \(\mu_s=0.40\); Q cannot remain at rest because its actual friction is below \(f_{s,\max}\).
Correct Answer: \(\mu_s=0.40\); Q can remain at rest with \(f=20\,\text{N}\); R cannot remain at rest.
Explanation: Trial P is at limiting equilibrium, so
\[
16=\mu_s(40).
\]
Therefore,
\[
\mu_s=0.40.
\]
For Trial Q, the maximum available static friction is
\[
f_{s,\max}=0.40(60)=24\,\text{N}.
\]
Only \(20\,\text{N}\) is required, so the block can remain at rest with
\[
f_s=20\,\text{N}.
\]
It is not at limiting equilibrium because \(20\lt 24\). For Trial R,
\[
f_{s,\max}=0.40(30)=12\,\text{N}.
\]
The required \(14\,\text{N}\) exceeds this value. Therefore the recorded rest state in R is incompatible with the stated model. The comparison also demonstrates that a given surface pair does not imply one fixed friction force independent of normal load.
407. A \(2\,\text{kg}\) block is placed on a rough plane whose inclination can be slowly increased. The coefficients are \(\mu_s=0.75\) and \(\mu_k=0.50\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). Which statement is correct?
ⓐ. At \(30^\circ\): \(f_s=10\,\text{N}\) up the plane; at \(37^\circ\): limiting equilibrium; at \(45^\circ\): \(a=2.5\sqrt2\,\text{m s}^{-2}\) down the plane.
ⓑ. At \(30^\circ\): \(f_s=f_{s,\max}\); at \(37^\circ\): \(f_s=0\); at \(45^\circ\): \(a=2.5\sqrt2\,\text{m s}^{-2}\) up the plane.
ⓒ. At \(30^\circ\): limiting equilibrium; before \(37^\circ\): sliding begins because \(\mu_k\lt \mu_s\); at \(45^\circ\): \(a=0\).
ⓓ. At \(30^\circ\): \(f_s=10\,\text{N}\) down the plane; at \(37^\circ\): limiting equilibrium; at \(45^\circ\): \(a=0\).
Correct Answer: At \(30^\circ\): \(f_s=10\,\text{N}\) up the plane; at \(37^\circ\): limiting equilibrium; at \(45^\circ\): \(a=2.5\sqrt2\,\text{m s}^{-2}\) down the plane.
Explanation: At \(30^\circ\), the downslope component of weight is
\[
mg\sin30^\circ=2(10)\left(\frac12\right)=10\,\text{N}.
\]
Static friction therefore takes the actual value \(10\,\text{N}\) up the plane. At \(37^\circ\),
\[
mg\sin37^\circ=12\,\text{N},
\]
while
\[
N=mg\cos37^\circ=16\,\text{N}.
\]
Thus,
\[
f_{s,\max}=0.75(16)=12\,\text{N},
\]
so the block is exactly at limiting equilibrium. Once actual sliding occurs at \(45^\circ\), kinetic friction must be used. The acceleration is
\[
a=g(\sin45^\circ-\mu_k\cos45^\circ).
\]
Hence,
\[
a=10\left(\frac1{\sqrt2}-\frac{0.5}{\sqrt2}\right)
=\frac5{\sqrt2}
=2.5\sqrt2\,\text{m s}^{-2}.
\]
The static threshold must therefore be checked before selecting the kinetic-friction model.
408. A \(5\,\text{kg}\) block is pressed against a vertical wall by a force \(P\) directed \(37^\circ\) above the horizontal towards the wall. The coefficient of static friction is \(0.50\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). For what range of \(P\) can the block remain at rest? Also, what is the friction direction when \(P=100\,\text{N}\)?
ⓐ. \(0\le P\le250\,\text{N}\); friction is zero at \(P=100\,\text{N}\)
ⓑ. \(50\,\text{N}\le P\le100\,\text{N}\); friction is \(10\,\text{N}\) upward
ⓒ. \(50\,\text{N}\le P\le250\,\text{N}\); friction is \(10\,\text{N}\) downward at \(P=100\,\text{N}\)
ⓓ. \(62.5\,\text{N}\le P\le125\,\text{N}\); friction is \(20\,\text{N}\) downward
Correct Answer: \(50\,\text{N}\le P\le250\,\text{N}\); friction is \(10\,\text{N}\) downward at \(P=100\,\text{N}\)
Explanation: The force component pressing the block into the wall is
\[
P\cos37^\circ=0.8P,
\]
so
\[
N=0.8P.
\]
The maximum static friction is therefore
\[
f_{s,\max}=0.50(0.8P)=0.4P.
\]
The upward component of the applied force is
\[
0.6P.
\]
Vertical equilibrium requires friction
\[
f=50-0.6P,
\]
where a negative value means downward friction. Thus,
\[
|50-0.6P|\le0.4P.
\]
One side gives
\[
P\ge50\,\text{N},
\]
while the other gives
\[
P\le250\,\text{N}.
\]
At \(P=100\,\text{N}\),
\[
f=50-60=-10\,\text{N},
\]
so friction has magnitude \(10\,\text{N}\) and acts downward because the applied force alone would tend to move the block upward.
409. A block remains at rest relative to a horizontal conveyor belt throughout two successive intervals. During the first interval the belt accelerates towards the right. During the second interval the belt and block are still moving towards the right, but their acceleration is towards the left. Consider the following statements.
Statement I: Static friction on the block may act towards the right during the first interval.
Statement II: Static friction on the block may act towards the left during the second interval even though the block's velocity is still towards the right.
Statement III: In both intervals the contact surfaces have zero relative sliding while static friction is non-zero.
Statement IV: Static friction must always point opposite to the block's instantaneous velocity.
Which option is correct?
ⓐ. II and IV only
ⓑ. I and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: The block must acquire the same acceleration as the belt if no relative slipping occurs. During the first interval, a rightward resultant is needed on the block, so static friction can act rightward. During the second interval, the velocity may still be rightward while the acceleration is leftward. Static friction can then act leftward to decelerate the block together with the belt. Thus friction direction is controlled by the tendency of relative slipping at the contact, not by the translational velocity alone. Since the block remains at rest relative to the belt, the contact is in the static-friction regime in both intervals. Non-zero static friction is entirely compatible with zero relative sliding. Hence Statements I, II and III are correct. Statement IV is the misconception that friction must always oppose the body's velocity relative to the ground.
410. A \(5\,\text{kg}\) block rests on a rough horizontal surface. A horizontal force of \(20\,\text{N}\) places it just at impending motion. Take \(g=10\,\text{m s}^{-2}\). The force is then reduced to \(18\,\text{N}\), but an additional vertical force of \(20\,\text{N}\) upward is applied while contact with the floor is maintained. What happens?
ⓐ. The block remains at rest with \(18\,\text{N}\) static friction because \(18\lt 20\).
ⓑ. The block starts sliding because the upward force reduces the normal reaction and the new maximum static friction is only \(12\,\text{N}\).
ⓒ. The block remains at limiting equilibrium because the coefficient of friction changes when the vertical force is applied.
ⓓ. The block starts sliding because the upward force directly produces a horizontal component.
Correct Answer: The block starts sliding because the upward force reduces the normal reaction and the new maximum static friction is only \(12\,\text{N}\).
Explanation: In the original limiting condition,
\[
20=\mu_s(50).
\]
Therefore,
\[
\mu_s=0.40.
\]
After the upward \(20\,\text{N}\) force is applied, the normal reaction becomes
\[
N=50-20=30\,\text{N}.
\]
The surface pair is unchanged, so the coefficient remains
\[
\mu_s=0.40.
\]
Hence the new maximum static friction is
\[
f_{s,\max}=0.40(30)=12\,\text{N}.
\]
The horizontal force now requires \(18\,\text{N}\) of opposing friction to maintain rest. Since
\[
18\gt 12,
\]
static friction is insufficient and sliding begins. The fact that \(18\,\text{N}\) is smaller than the old \(20\,\text{N}\) threshold is irrelevant after the normal reaction has changed.
411. A \(60\,\text{kg}\) person stands on a horizontal floor with coefficient of static friction \(0.40\) and pushes horizontally on a rigid wall with force \(250\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\). In a second trial the person carries a \(20\,\text{kg}\) backpack rigidly while exerting the same wall force. Which statement is correct?
ⓐ. Without backpack: equilibrium possible; with backpack: equilibrium possible and floor friction \(250\,\text{N}\).
ⓑ. Without backpack: equilibrium impossible; with backpack: equilibrium possible and floor friction \(250\,\text{N}\).
ⓒ. Without backpack: equilibrium impossible; with backpack: equilibrium impossible because \(\mu_s\) is unchanged.
ⓓ. Without backpack: equilibrium impossible; with backpack: equilibrium possible and floor friction \(320\,\text{N}\).
Correct Answer: Without backpack: equilibrium impossible; with backpack: equilibrium possible and floor friction \(250\,\text{N}\).
Explanation: When the person pushes the wall with \(250\,\text{N}\), the wall simultaneously pushes the person with \(250\,\text{N}\) in the opposite direction. For horizontal equilibrium, the floor must provide \(250\,\text{N}\) of static friction opposite to the wall's force on the person. Without the backpack,
\[
N=60(10)=600\,\text{N}.
\]
Thus,
\[
f_{s,\max}=0.40(600)=240\,\text{N}.
\]
Since \(250\gt 240\), static equilibrium is impossible. With the \(20\,\text{kg}\) backpack,
\[
N=80(10)=800\,\text{N},
\]
and
\[
f_{s,\max}=0.40(800)=320\,\text{N}.
\]
Only \(250\,\text{N}\) is required, so equilibrium is now possible with actual friction \(250\,\text{N}\), not \(320\,\text{N}\). The floor-friction partner is the equal-and-opposite force exerted by the person on the floor.
412. A block slides on a rough plane inclined at \(37^\circ\) above the horizontal. The coefficient of kinetic friction is \(0.25\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). Compare the acceleration while the block is sliding upward with the acceleration after it reverses and slides downward under gravity alone.
ⓐ. While sliding upward: \(4\,\text{m s}^{-2}\) down the plane; while sliding downward: \(8\,\text{m s}^{-2}\) down the plane.
ⓑ. While sliding upward: \(4\,\text{m s}^{-2}\) down the plane; while sliding downward: \(4\,\text{m s}^{-2}\) down the plane.
ⓒ. While sliding upward: \(8\,\text{m s}^{-2}\) down the plane; while sliding downward: \(4\,\text{m s}^{-2}\) down the plane.
ⓓ. While sliding upward: \(8\,\text{m s}^{-2}\) up the plane; while sliding downward: \(4\,\text{m s}^{-2}\) down the plane.
Correct Answer: While sliding upward: \(8\,\text{m s}^{-2}\) down the plane; while sliding downward: \(4\,\text{m s}^{-2}\) down the plane.
Explanation: The normal reaction is
\[
N=mg\cos37^\circ.
\]
Hence the kinetic-friction magnitude is
\[
f_k=\mu_kmg\cos37^\circ.
\]
While the block slides upward, both gravity's tangential component and kinetic friction act down the plane. Therefore,
\[
a_{\text{up-motion}}
=g(\sin37^\circ+\mu_k\cos37^\circ).
\]
Thus,
\[
a_{\text{up-motion}}
=10(0.6+0.25\times0.8)
=8\,\text{m s}^{-2}
\]
down the plane. After the block reverses and slides downward, kinetic friction reverses direction and acts up the plane. Hence,
\[
a_{\text{down-motion}}
=g(\sin37^\circ-\mu_k\cos37^\circ)
=10(0.6-0.2)
=4\,\text{m s}^{-2}
\]
down the plane. Friction opposes actual relative sliding, so one memorized incline formula cannot be used for both directions.
413. A \(5\,\text{kg}\) block is already sliding towards the right on a horizontal rough surface with \(\mu_k=0.25\). A \(20\,\text{N}\) force is applied at \(37^\circ\) to the horizontal towards the right. Compare Case I, where it pulls upward, with Case II, where it pushes downward. Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). What are the accelerations?
ⓐ. Case I: \(3.2\,\text{m s}^{-2}\) right; Case II: \(3.2\,\text{m s}^{-2}\) right
ⓑ. Case I: \(0.1\,\text{m s}^{-2}\) right; Case II: \(1.3\,\text{m s}^{-2}\) right
ⓒ. Case I: \(1.3\,\text{m s}^{-2}\) right; Case II: \(0.1\,\text{m s}^{-2}\) right
ⓓ. Case I: \(1.3\,\text{m s}^{-2}\) left; Case II: \(0.1\,\text{m s}^{-2}\) left
Correct Answer: Case I: \(1.3\,\text{m s}^{-2}\) right; Case II: \(0.1\,\text{m s}^{-2}\) right
Explanation: The horizontal component of the applied force is
\[
F_x=20(0.8)=16\,\text{N}
\]
in both cases. In Case I, the upward component reduces the normal reaction:
\[
N_1=50-20(0.6)=38\,\text{N}.
\]
Thus,
\[
f_{k1}=0.25(38)=9.5\,\text{N}.
\]
Since the block slides rightward, friction acts leftward, so
\[
F_{\text{net},1}=16-9.5=6.5\,\text{N}.
\]
Hence,
\[
a_1=\frac{6.5}{5}=1.3\,\text{m s}^{-2}.
\]
In Case II,
\[
N_2=50+12=62\,\text{N},
\]
so
\[
f_{k2}=0.25(62)=15.5\,\text{N}.
\]
Therefore,
\[
a_2=\frac{16-15.5}{5}=0.1\,\text{m s}^{-2}
\]
rightward. The downward push nearly cancels its own horizontal driving effect because it simultaneously increases kinetic friction.
414. A \(2\,\text{kg}\) block \(A\) is sliding towards the right relative to a \(3\,\text{kg}\) block \(B\). The floor below \(B\) is smooth, and the coefficient of kinetic friction between \(A\) and \(B\) is \(0.30\). Take \(g=10\,\text{m s}^{-2}\). Which statement is correct while relative sliding continues?
ⓐ. On \(A\): \(6\,\text{N}\) leftward, \(a_A=3\,\text{m s}^{-2}\) leftward; on \(B\): \(6\,\text{N}\) rightward, \(a_B=2\,\text{m s}^{-2}\) rightward.
ⓑ. On \(A\): \(6\,\text{N}\) leftward, \(a_A=3\,\text{m s}^{-2}\) leftward; on \(B\): \(6\,\text{N}\) leftward, \(a_B=2\,\text{m s}^{-2}\) leftward.
ⓒ. On \(A\): \(9\,\text{N}\) leftward, \(a_A=4.5\,\text{m s}^{-2}\) leftward; on \(B\): \(9\,\text{N}\) rightward, \(a_B=3\,\text{m s}^{-2}\) rightward.
ⓓ. On both blocks: kinetic friction is zero because the floor under \(B\) is smooth.
Correct Answer: On \(A\): \(6\,\text{N}\) leftward, \(a_A=3\,\text{m s}^{-2}\) leftward; on \(B\): \(6\,\text{N}\) rightward, \(a_B=2\,\text{m s}^{-2}\) rightward.
Explanation: The relevant sliding occurs at the \(A-B\) interface, not at the floor. The normal reaction between the blocks is
\[
N=m_Ag=2(10)=20\,\text{N}.
\]
Therefore,
\[
f_k=\mu_kN=0.30(20)=6\,\text{N}.
\]
Since \(A\) slides rightward relative to \(B\), kinetic friction on \(A\) acts leftward. By Newton's Third Law, the friction exerted by \(A\) on \(B\) is \(6\,\text{N}\) rightward. Thus,
\[
a_A=\frac{6}{2}=3\,\text{m s}^{-2}
\]
leftward. For \(B\),
\[
a_B=\frac{6}{3}=2\,\text{m s}^{-2}
\]
rightward. The two friction forces are equal and opposite, but the accelerations differ because the masses differ. The direction must be determined from relative sliding at the contact.
415. A \(2\,\text{kg}\) block is placed on a horizontal conveyor belt moving steadily towards the right at \(8\,\text{m s}^{-1}\). At the instant considered, the block is already moving towards the right at \(2\,\text{m s}^{-1}\), so it is sliding relative to the belt. The coefficient of kinetic friction is \(0.20\), and \(g=10\,\text{m s}^{-2}\). Neglect other horizontal forces. Which statement is correct during the sliding phase?
ⓐ. Friction \(4\,\text{N}\) leftward; \(a=2\,\text{m s}^{-2}\) leftward; after \(2\,\text{s}\), \(v=2\,\text{m s}^{-1}\) leftward and sliding continues.
ⓑ. Friction \(4\,\text{N}\) rightward; \(a=2\,\text{m s}^{-2}\) rightward; after \(2\,\text{s}\), \(v=6\,\text{m s}^{-1}\) and sliding continues.
ⓒ. Friction \(8\,\text{N}\) rightward; \(a=4\,\text{m s}^{-2}\) rightward; the block instantly acquires the belt speed.
ⓓ. Friction is zero; \(a=0\); the block keeps moving at \(2\,\text{m s}^{-1}\) because the belt speed is constant.
Correct Answer: Friction \(4\,\text{N}\) rightward; \(a=2\,\text{m s}^{-2}\) rightward; after \(2\,\text{s}\), \(v=6\,\text{m s}^{-1}\) and sliding continues.
Explanation: The belt moves rightward faster than the block. Therefore the block slides leftward relative to the belt surface, even though both velocities relative to the ground are rightward. Kinetic friction opposes this relative sliding, so it acts on the block towards the right. The normal reaction is
\[
N=mg=2(10)=20\,\text{N}.
\]
Thus,
\[
f_k=0.20(20)=4\,\text{N}.
\]
The block's acceleration is
\[
a=\frac42=2\,\text{m s}^{-2}
\]
rightward. After \(2\,\text{s}\),
\[
v=2+2(2)=6\,\text{m s}^{-1}.
\]
This is still less than the belt speed \(8\,\text{m s}^{-1}\), so relative sliding persists at that instant. The friction regime must be reconsidered only when the relative sliding actually ceases.
416. A \(4\,\text{kg}\) block is on a rough plane inclined at \(37^\circ\) above the horizontal. A constant horizontal force of \(20\,\text{N}\) acts towards the rising side of the plane. The coefficient of kinetic friction is \(0.25\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). Compare the acceleration when the block is sliding upward with that when it is sliding downward.
ⓐ. Sliding upward: \(4.75\,\text{m s}^{-2}\) down the plane; sliding downward: \(0.75\,\text{m s}^{-2}\) up the plane
ⓑ. Sliding upward: \(0.75\,\text{m s}^{-2}\) down; sliding downward: \(4.75\,\text{m s}^{-2}\) down
ⓒ. Both directions give \(2\,\text{m s}^{-2}\) down the plane.
ⓓ. Sliding upward: \(4.75\,\text{m s}^{-2}\) up; sliding downward: \(0.75\,\text{m s}^{-2}\) down
Correct Answer: Sliding upward: \(4.75\,\text{m s}^{-2}\) down the plane; sliding downward: \(0.75\,\text{m s}^{-2}\) up the plane
Explanation: The horizontal force has an up-slope component
\[
20\cos37^\circ=16\,\text{N}.
\]
It also presses the block into the plane by
\[
20\sin37^\circ=12\,\text{N}.
\]
Therefore,
\[
N=mg\cos37^\circ+12=32+12=44\,\text{N}.
\]
The kinetic-friction magnitude is
\[
f_k=0.25(44)=11\,\text{N}.
\]
Gravity contributes
\[
mg\sin37^\circ=24\,\text{N}
\]
down the plane. While the block slides upward, friction also acts down the plane, so the net downslope force is
\[
24+11-16=19\,\text{N}.
\]
Hence,
\[
a=\frac{19}{4}=4.75\,\text{m s}^{-2}
\]
down the plane. While the block slides downward, friction acts upward, producing
\[
16+11-24=3\,\text{N}
\]
up the plane. Thus,
\[
a=\frac34=0.75\,\text{m s}^{-2}
\]
up the plane.
417. A \(2\,\text{kg}\) block \(A\) rests on a \(3\,\text{kg}\) block \(B\), and the floor below \(B\) is smooth. The coefficients between \(A\) and \(B\) are \(\mu_s=0.40\) and \(\mu_k=0.30\). A horizontal force acts on \(A\). In Stage I its magnitude is \(10\,\text{N}\); in Stage II it is suddenly increased to \(30\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. The blocks move together in both stages with accelerations \(2\,\text{m s}^{-2}\) and \(6\,\text{m s}^{-2}\).
ⓑ. In Stage I they move together at \(2\,\text{m s}^{-2}\); in Stage II they slip, with \(a_A=12\,\text{m s}^{-2}\) and \(a_B=2\,\text{m s}^{-2}\).
ⓒ. They slip in Stage I but move together in Stage II.
ⓓ. In Stage II the kinetic friction is \(8\,\text{N}\), so \(a_A=11\,\text{m s}^{-2}\).
Correct Answer: In Stage I they move together at \(2\,\text{m s}^{-2}\); in Stage II they slip, with \(a_A=12\,\text{m s}^{-2}\) and \(a_B=2\,\text{m s}^{-2}\).
Explanation: In Stage I, assuming no slip, the common acceleration is
\[
a_1=\frac{10}{2+3}=2\,\text{m s}^{-2}.
\]
Block \(B\) must be accelerated by friction alone, so the required friction is
\[
f_1=3(2)=6\,\text{N}.
\]
The maximum static friction is
\[
f_{s,\max}=0.40(2)(10)=8\,\text{N}.
\]
Since \(6\lt 8\), the no-slip assumption is valid. In Stage II, a common acceleration would be
\[
a=\frac{30}{5}=6\,\text{m s}^{-2}.
\]
That would require
\[
f=3(6)=18\,\text{N}
\]
on block \(B\), which exceeds the \(8\,\text{N}\) static limit. Sliding therefore begins and the friction changes to
\[
f_k=0.30(20)=6\,\text{N}.
\]
For \(A\),
\[
a_A=\frac{30-6}{2}=12\,\text{m s}^{-2},
\]
while for \(B\),
\[
a_B=\frac63=2\,\text{m s}^{-2}.
\]
418. A \(4\,\text{kg}\) block rests on a rough plane inclined at \(37^\circ\) above the horizontal. The coefficient of static friction is \(0.50\). A force \(P\) is applied at \(37^\circ\) above the plane, directed up the slope and away from the surface. Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). For what range of \(P\) can the block remain in equilibrium while maintaining contact with the plane?
ⓐ. \(16\,\text{N}\le P\le\dfrac{400}{11}\,\text{N}\)
ⓑ. \(0\le P\le30\,\text{N}\)
ⓒ. \(24\,\text{N}\le P\le40\,\text{N}\)
ⓓ. \(16\,\text{N}\le P\le30\,\text{N}\)
Correct Answer: \(16\,\text{N}\le P\le\dfrac{400}{11}\,\text{N}\)
Explanation: The block's weight is
\[
mg=40\,\text{N}.
\]
Its component down the plane is
\[
40\sin37^\circ=24\,\text{N},
\]
while the inward normal component is
\[
40\cos37^\circ=32\,\text{N}.
\]
The applied force has an up-slope component
\[
P\cos37^\circ=0.8P
\]
and an outward normal component
\[
P\sin37^\circ=0.6P.
\]
Hence,
\[
N=32-0.6P.
\]
The friction required for equilibrium is
\[
f=24-0.8P,
\]
where the sign determines its direction. Static equilibrium requires
\[
|24-0.8P|\le0.5(32-0.6P)=16-0.3P.
\]
From \(24-0.8P\le16-0.3P\),
\[
P\ge16\,\text{N}.
\]
From \(-24+0.8P\le16-0.3P\),
\[
P\le\frac{400}{11}\,\text{N}.
\]
This upper value is below the contact-loss condition \(P=32/0.6\), so contact is maintained throughout the allowed range.
419. A \(5\,\text{kg}\) block is pressed against a vertical rough wall by a force \(P\) directed at \(37^\circ\) below the horizontal and towards the wall. The coefficient of static friction between the block and wall is \(0.50\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). Which statement is correct?
ⓐ. The block can be held at rest if \(P\ge125\,\text{N}\).
ⓑ. The block can be held at rest only for \(50\,\text{N}\le P\le250\,\text{N}\).
ⓒ. No finite positive value of \(P\) can hold the block at rest under the stated force direction.
ⓓ. The minimum required force is \(250\,\text{N}\).
Correct Answer: No finite positive value of \(P\) can hold the block at rest under the stated force direction.
Explanation: The component of \(P\) pressing the block into the wall is
\[
P\cos37^\circ=0.8P.
\]
Thus the normal reaction is
\[
N=0.8P.
\]
The maximum available static friction is therefore
\[
f_{s,\max}=0.50(0.8P)=0.4P.
\]
However, the applied force also has a downward component
\[
P\sin37^\circ=0.6P.
\]
Together with the weight, the total downward force that friction would have to balance is
\[
50+0.6P.
\]
Equilibrium would require
\[
0.4P\ge50+0.6P.
\]
This implies
\[
-0.2P\ge50,
\]
which is impossible for any positive \(P\). Increasing \(P\) increases the available friction, but it increases the additional downward component even faster. Therefore no finite positive \(P\) can produce equilibrium in this geometry.
420. A \(2\,\text{kg}\) block rests without slipping on the horizontal floor of a truck. The coefficient of static friction between the block and truck is \(0.50\). An external force of \(10\,\text{N}\) acts on the block at \(37^\circ\) above the horizontal towards the right. The truck accelerates towards the right with acceleration \(a\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\) and \(\cos37^\circ=0.8\). For what range of \(a\) can the block remain at rest relative to the truck?
ⓐ. \(0\le a\le7\,\text{m s}^{-2}\)
ⓑ. \(0.5\,\text{m s}^{-2}\le a\le7.5\,\text{m s}^{-2}\), with friction reversing at \(a=4\,\text{m s}^{-2}\)
ⓒ. \(1\,\text{m s}^{-2}\le a\le8\,\text{m s}^{-2}\), with friction always rightward
ⓓ. \(0.5\,\text{m s}^{-2}\le a\le4\,\text{m s}^{-2}\), with friction always leftward
Correct Answer: \(0.5\,\text{m s}^{-2}\le a\le7.5\,\text{m s}^{-2}\), with friction reversing at \(a=4\,\text{m s}^{-2}\)
Explanation: The upward component of the applied force is
\[
10\sin37^\circ=6\,\text{N}.
\]
Therefore the normal reaction is
\[
N=20-6=14\,\text{N}.
\]
The maximum static friction is
\[
f_{s,\max}=0.50(14)=7\,\text{N}.
\]
The rightward component of the external force is
\[
10\cos37^\circ=8\,\text{N}.
\]
If the block moves with the truck, its horizontal acceleration must be \(a\). Taking rightward friction as positive,
\[
8+f=2a.
\]
Thus,
\[
f=2a-8.
\]
For no slipping,
\[
|2a-8|\le7.
\]
Hence,
\[
0.5\le a\le7.5\,\text{m s}^{-2}.
\]
Friction is zero at \(a=4\,\text{m s}^{-2}\), acts leftward below this acceleration, and rightward above it.
