401. A uniform horizontal beam of length \(6\,\text{m}\) and weight \(200\,\text{N}\) is supported vertically at its two ends A and B. A \(300\,\text{N}\) point load is placed \(1\,\text{m}\) from A. If the beam is in equilibrium, the upward reactions \(R_A\) and \(R_B\) are
ⓐ. \(400\,\text{N}\) and \(100\,\text{N}\)
ⓑ. \(150\,\text{N}\) and \(350\,\text{N}\)
ⓒ. \(250\,\text{N}\) and \(250\,\text{N}\)
ⓓ. \(350\,\text{N}\) and \(150\,\text{N}\)
Correct Answer: \(350\,\text{N}\) and \(150\,\text{N}\)
Explanation: The beam is uniform, so its own weight acts at its midpoint, \(3\,\text{m}\) from A.
\(\textbf{Choose A as the moment pivot:}\)
The reaction \(R_A\) then contributes zero moment.
Torque balance gives
\[
R_B(6)=200(3)+300(1)
\]
Therefore,
\[
6R_B=600+300=900
\]
Hence,
\[
R_B=150\,\text{N}
\]
\(\textbf{Apply vertical force balance:}\)
\[
R_A+R_B=200+300
\]
Thus,
\[
R_A+150=500
\]
Therefore,
\[
R_A=350\,\text{N}
\]
The support nearer the additional \(300\,\text{N}\) load carries the larger reaction, which is physically consistent.
\(\textbf{Final answer:}\)
\[
\boxed{R_A=350\,\text{N},\qquad R_B=150\,\text{N}}
\]
402. A non-uniform horizontal beam of length \(4\,\text{m}\) and total weight \(200\,\text{N}\) is supported at its ends A and B. In equilibrium, the upward reactions are \(80\,\text{N}\) at A and \(120\,\text{N}\) at B. The centre of gravity of the beam is located at a distance
ⓐ. \(1.6\,\text{m}\) from A
ⓑ. \(2.0\,\text{m}\) from A
ⓒ. \(2.4\,\text{m}\) from A
ⓓ. \(3.0\,\text{m}\) from A
Correct Answer: \(2.4\,\text{m}\) from A
Explanation: The reactions add to
\[
80+120=200\,\text{N},
\]
so the vertical force-balance condition is already satisfied.
Let the beam's centre of gravity be at distance \(x\) from A.
Its entire weight may be treated as acting vertically downward through that point.
\(\textbf{Take moments about A:}\)
The reaction at A produces zero moment.
Therefore,
\[
120(4)=200x
\]
Hence,
\[
480=200x
\]
Thus,
\[
x=2.4\,\text{m}
\]
The greater reaction at B indicates that the centre of gravity lies closer to B than to A.
Because the beam is non-uniform, its centre of gravity need not be at the geometric midpoint.
\(\textbf{Final answer:}\)
\[
x_{\rm CG}=2.4\,\text{m from A}
\]
403. A horizontal beam of length \(8\,\text{m}\) is supported at its ends A and B. Its own weight is \(100\,\text{N}\) and acts at the midpoint. An additional load \(W\) acts downward \(6\,\text{m}\) from A. In equilibrium the upward reaction at A is \(150\,\text{N}\). The load \(W\) and the reaction at B are
ⓐ. \(300\,\text{N}\) and \(250\,\text{N}\)
ⓑ. \(400\,\text{N}\) and \(350\,\text{N}\)
ⓒ. \(500\,\text{N}\) and \(450\,\text{N}\)
ⓓ. \(200\,\text{N}\) and \(150\,\text{N}\)
Correct Answer: \(400\,\text{N}\) and \(350\,\text{N}\)
Explanation: Let the upward reaction at B be \(R_B\).
\(\textbf{Vertical force balance:}\)
\[
150+R_B=100+W
\]
Therefore,
\[
R_B=W-50. \tag{1}
\]
\(\textbf{Moment balance about A:}\)
The reaction at A has zero moment about A.
Thus,
\[
R_B(8)=100(4)+W(6)
\]
Substituting Eq. (1),
\[
8(W-50)=400+6W
\]
Hence,
\[
8W-400=400+6W
\]
Therefore,
\[
2W=800,
\]
so
\[
W=400\,\text{N}
\]
Then,
\[
R_B=400-50=350\,\text{N}
\]
Both the force and torque equilibrium conditions are satisfied by these values.
\(\textbf{Final answer:}\)
\[
\boxed{W=400\,\text{N},\qquad R_B=350\,\text{N}}
\]
404. A uniform horizontal rod of length \(L\) and weight \(W\) is hinged at its left end. A load \(2W\) is attached at a distance \(3L/4\) from the hinge. The right end of the rod is supported by a light string making an angle \(30^\circ\) above the rod. For equilibrium, the tension in the string and the magnitude of the hinge reaction are
ⓐ. \(4W\) and \(\sqrt{13}\,W\)
ⓑ. \(2W\) and \(\sqrt5\,W\)
ⓒ. \(4W\) and \(3W\)
ⓓ. \(2\sqrt3\,W\) and \(\sqrt{13}\,W\)
Correct Answer: \(4W\) and \(\sqrt{13}\,W\)
Explanation: The hinge force is unknown, so taking moments about the hinge eliminates both of its components.
\(\textbf{Moment balance about the hinge:}\)
The vertical component of the string tension is
\[
T\sin30^\circ=\frac T2
\]
Its anticlockwise moment is
\[
\frac T2L
\]
The rod weight produces the clockwise moment
\[
W\left(\frac L2\right)
\]
The attached load produces
\[
2W\left(\frac{3L}{4}\right)
\]
Equilibrium therefore requires
\[
\frac T2L=\frac{WL}{2}+\frac{3WL}{2}
\]
Hence,
\[
\frac T2=2W,
\]
so
\[
T=4W
\]
\(\textbf{Resolve the hinge reaction:}\)
The string components are
\[
T_x=4W\cos30^\circ=2\sqrt3\,W,
\]
\[
T_y=4W\sin30^\circ=2W
\]
The total downward load is \(3W\), so the hinge must provide an upward component
\[
H_y=3W-2W=W
\]
Its horizontal component must balance the string:
\[
H_x=2\sqrt3\,W
\]
Therefore,
\[
H=\sqrt{H_x^2+H_y^2}
=\sqrt{12W^2+W^2}
=\sqrt{13}\,W
\]
\(\textbf{Final answer:}\)
\[
\boxed{T=4W,\qquad H=\sqrt{13}\,W}
\]
405. A uniform plank of length \(4\,\text{m}\) and weight \(200\,\text{N}\) rests horizontally on two narrow supports located \(0.5\,\text{m}\) and \(3.5\,\text{m}\) from its left end. A person of weight \(600\,\text{N}\), initially near the centre, walks slowly towards the right end. At the instant the plank is just about to lose contact with the left support, the person's position from the left end and the reaction at the right support are
ⓐ. \(3.5\,\text{m}\) and \(600\,\text{N}\)
ⓑ. \(4.0\,\text{m}\) and \(800\,\text{N}\)
ⓒ. \(3.75\,\text{m}\) and \(800\,\text{N}\)
ⓓ. \(4.0\,\text{m}\) and \(600\,\text{N}\)
Correct Answer: \(4.0\,\text{m}\) and \(800\,\text{N}\)
Explanation: At impending loss of contact, the reaction at the left support becomes zero.
The plank is then on the verge of turning about the right support.
\(\textbf{Use the right support as the moment point:}\)
The plank's weight acts at its centre, \(2.0\,\text{m}\) from the left end.
Its distance from the right support is
\[
3.5-2.0=1.5\,\text{m}
\]
Let the person's position be \(x\).
For \(x\gt 3.5\,\text{m}\), the person's moment opposes that of the plank's weight.
At the tipping threshold,
\[
600(x-3.5)=200(1.5)
\]
Thus,
\[
600(x-3.5)=300
\]
Hence,
\[
x-3.5=0.5,
\]
so
\[
x=4.0\,\text{m}
\]
\(\textbf{Force balance at the threshold:}\)
Only the right support provides an upward reaction.
Therefore,
\[
R_B=600+200=800\,\text{N}
\]
\(\textbf{Final answer:}\)
\[
\boxed{x=4.0\,\text{m},\qquad R_B=800\,\text{N}}
\]
406. A non-uniform horizontal rod of length \(5\,\text{m}\) and weight \(400\,\text{N}\) is supported at its ends A and B. A \(400\,\text{N}\) load is placed \(1\,\text{m}\) from A. In equilibrium, the reactions at A and B are \(500\,\text{N}\) and \(300\,\text{N}\), respectively. The position of the rod's centre of gravity from A and the reaction at B after the additional load is removed are
ⓐ. \(2.25\,\text{m}\) and \(180\,\text{N}\)
ⓑ. \(2.75\,\text{m}\) and \(220\,\text{N}\)
ⓒ. \(3.00\,\text{m}\) and \(240\,\text{N}\)
ⓓ. \(2.75\,\text{m}\) and \(300\,\text{N}\)
Correct Answer: \(2.75\,\text{m}\) and \(220\,\text{N}\)
Explanation: Let the rod's centre of gravity be at distance \(x\) from A.
\(\textbf{Infer the centre of gravity from the loaded condition:}\)
Taking moments about A,
\[
300(5)=400x+400(1)
\]
Therefore,
\[
1500=400x+400
\]
Hence,
\[
400x=1100,
\]
so
\[
x=2.75\,\text{m}
\]
\(\textbf{Remove the external load:}\)
Let the new reaction at B be \(R_B'\).
Moment balance about A now gives
\[
R_B'(5)=400(2.75)
\]
Thus,
\[
5R_B'=1100
\]
Therefore,
\[
R_B'=220\,\text{N}
\]
The corresponding reaction at A would be \(180\,\text{N}\), and the two reactions correctly add to the rod's weight.
\(\textbf{Final answer:}\)
\[
\boxed{x_{\rm CG}=2.75\,\text{m},\qquad R_B'=220\,\text{N}}
\]
407. A uniform ladder of weight \(W\) rests against a smooth vertical wall and a rough horizontal floor, making an angle \(60^\circ\) with the floor. A person of weight \(W\) stands at a point \(3L/4\) from the lower end along the ladder, where \(L\) is the ladder length. For equilibrium, the friction force at the floor and the normal reaction from the floor are
ⓐ. \(\displaystyle \frac{5W}{4\sqrt3}\) and \(2W\)
ⓑ. \(\displaystyle \frac{5W}{8}\) and \(W\)
ⓒ. \(\displaystyle \frac{3W}{2\sqrt3}\) and \(2W\)
ⓓ. \(\displaystyle \frac{5W}{4\sqrt3}\) and \(W\)
Correct Answer: \(\displaystyle \frac{5W}{4\sqrt3}\) and \(2W\)
Explanation: The smooth wall exerts only a horizontal normal reaction \(N_w\).
The floor provides an upward normal reaction and horizontal friction.
\(\textbf{Take moments about the lower end:}\)
The wall reaction has moment arm
\[
L\sin60^\circ
\]
The ladder's weight acts at \(L/2\), giving horizontal lever arm
\[
\frac L2\cos60^\circ
\]
The person's weight acts at \(3L/4\), giving horizontal lever arm
\[
\frac{3L}{4}\cos60^\circ
\]
Hence,
\[
N_wL\sin60^\circ
=W\frac L2\cos60^\circ
+W\frac{3L}{4}\cos60^\circ
\]
Using
\[
\sin60^\circ=\frac{\sqrt3}{2},
\qquad
\cos60^\circ=\frac12,
\]
we obtain
\[
N_w\frac{\sqrt3}{2}
=\frac W4+\frac{3W}{8}
=\frac{5W}{8}
\]
Therefore,
\[
N_w=\frac{5W}{4\sqrt3}
\]
\(\textbf{Horizontal and vertical force balance:}\)
The floor friction balances \(N_w\), so
\[
f=\frac{5W}{4\sqrt3}
\]
The wall force has no vertical component, hence
\[
N_{\rm floor}=W+W=2W
\]
\(\textbf{Final answer:}\)
\[
\boxed{f=\frac{5W}{4\sqrt3},\qquad N_{\rm floor}=2W}
\]
408. Consider the following statements for a rigid body in a plane.
Statement I: If the resultant external force is zero, the net external torque has the same value about every origin.
Statement II: If the resultant external force is zero and the net external torque is zero about one point, the force and torque equilibrium conditions are satisfied.
Statement III: If the net external torque is zero about one point, the resultant external force must also be zero.
Statement IV: A non-zero couple can be made to have zero moment merely by changing the origin.
Which combination is correct?
ⓐ. I and II only
ⓑ. I, II and III only
ⓒ. II and IV only
ⓓ. I, III and IV only
Correct Answer: I and II only
Explanation: Torque about different origins changes according to the resultant external force.
If
\[
\sum\vec F=0,
\]
shifting the origin does not change the net torque.
Therefore Statement I is correct.
If, in addition,
\[
\sum\vec\tau=0
\]
about one point, then both translational and rotational equilibrium conditions are fulfilled, so Statement II is correct.
Statement III is false because a single non-zero force whose line of action passes through the chosen point can have zero torque about that point.
Zero torque about one chosen point therefore does not by itself establish zero resultant force.
Statement IV is false because the moment of a pure couple is independent of the choice of origin.
A non-zero couple cannot be eliminated merely by moving the reference point.
Thus only Statements I and II are correct.
409. A horizontal rod of length \(L\) and weight \(W\) is hinged at its left end. A load \(2W\) hangs from its free end. A supporting string is attached at a point \(3L/4\) from the hinge and makes an angle \(60^\circ\) above the rod. In equilibrium, the tension in the string and the vertical component of the hinge reaction are
ⓐ. \(\displaystyle \frac{20W}{3\sqrt3}\) and \(\displaystyle \frac W3\) upward
ⓑ. \(\displaystyle \frac{20W}{3\sqrt3}\) and \(\displaystyle \frac W3\) downward
ⓒ. \(\displaystyle \frac{10W}{3\sqrt3}\) and \(\displaystyle \frac W3\) upward
ⓓ. \(\displaystyle \frac{10W}{\sqrt3}\) and \(W\) downward
Correct Answer: \(\displaystyle \frac{20W}{3\sqrt3}\) and \(\displaystyle \frac W3\) downward
Explanation: The hinge force is removed from the torque equation by taking moments about the hinge.
\(\textbf{Rotational equilibrium:}\)
The vertical component of the string tension is \(T\sin60^\circ\).
Its moment is
\[
T\sin60^\circ\left(\frac{3L}{4}\right)
\]
The clockwise moments of the weights are
\[
W\left(\frac L2\right)+2W(L)
\]
Therefore,
\[
T\sin60^\circ\frac{3L}{4}
=\frac{WL}{2}+2WL
\]
Hence,
\[
T\sin60^\circ\frac34=\frac52W
\]
Thus,
\[
T\sin60^\circ=\frac{10W}{3}
\]
Using \(\sin60^\circ=\sqrt3/2\),
\[
T=\frac{20W}{3\sqrt3}
\]
\(\textbf{Vertical force balance:}\)
Let upward hinge reaction be positive.
Then
\[
H_y+\frac{10W}{3}-3W=0
\]
Therefore,
\[
H_y=-\frac W3
\]
The negative sign means the hinge exerts a downward vertical force on the rod.
\(\textbf{Final answer:}\)
\[
T=\frac{20W}{3\sqrt3},
\]
with hinge vertical reaction \(W/3\) downward.
410. Assertion: If a rigid body is in equilibrium under exactly three coplanar, non-parallel forces, the three lines of action must be concurrent.
Reason: The intersection of the lines of action of any two forces may be chosen as a moment point; for zero net torque, the line of action of the third force must also pass through that point.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion
Explanation: Suppose the lines of action of two of the non-parallel forces intersect at point P.
Each of those two forces has zero moment about P because its line of action passes through P.
For rotational equilibrium, the total moment about P must also be zero.
Therefore the third force must have zero moment about P as well.
A non-zero force has zero moment about a point only when its line of action passes through that point.
Hence the third line of action must also pass through P.
Thus the three lines are concurrent.
This concurrency condition is necessary for equilibrium under exactly three coplanar non-parallel forces.
Force-vector balance must still also be satisfied.
The Reason therefore correctly explains the Assertion.
411. A rigid body is acted upon by three vertical forces along the \(x\)-axis: \(10\,\text{N}\) upward at \(x=2\,\text{m}\), \(6\,\text{N}\) downward at \(x=-1\,\text{m}\), and an unknown force \(F\) at \(x=4\,\text{m}\). An external couple \(C\) also acts. For complete equilibrium, \(F\) and \(C\) must be
ⓐ. \(4\,\text{N}\) downward and \(10\,\text{N m}\) anticlockwise
ⓑ. \(4\,\text{N}\) upward and \(10\,\text{N m}\) anticlockwise
ⓒ. \(16\,\text{N}\) downward and \(6\,\text{N m}\) clockwise
ⓓ. \(4\,\text{N}\) downward and \(10\,\text{N m}\) clockwise
Correct Answer: \(4\,\text{N}\) downward and \(10\,\text{N m}\) clockwise
Explanation: Complete equilibrium requires both zero resultant force and zero resultant torque.
\(\textbf{Vertical force balance:}\)
Taking upward as positive,
\[
10-6+F=0
\]
Therefore,
\[
F=-4\,\text{N}
\]
Thus the unknown force is \(4\,\text{N}\) downward.
\(\textbf{Torque of the three forces about the origin:}\)
For a vertical force,
\[
\tau_z=xF_y
\]
Hence,
\[
\tau_1=(2)(10)=20\,\text{N m},
\]
\[
\tau_2=(-1)(-6)=6\,\text{N m},
\]
and
\[
\tau_3=(4)(-4)=-16\,\text{N m}
\]
Therefore,
\[
\tau_{\rm forces}=20+6-16=10\,\text{N m}
\]
anticlockwise.
The required couple must cancel this:
\[
C=-10\,\text{N m}
\]
Hence it is \(10\,\text{N m}\) clockwise.
\(\textbf{Final answer:}\) \(4\,\text{N}\) downward and a \(10\,\text{N m}\) clockwise couple.
412. A uniform metre rule of weight \(W\) is balanced horizontally on a pivot at the \(40\,\text{cm}\) mark. A load \(2W\) hangs at the \(20\,\text{cm}\) mark and an unknown load \(P\) hangs at the \(80\,\text{cm}\) mark. The load \(P\) and the upward force exerted by the pivot are
ⓐ. \(\displaystyle \frac{W}{2}\) and \(\displaystyle \frac{7W}{2}\)
ⓑ. \(\displaystyle \frac{3W}{4}\) and \(\displaystyle \frac{15W}{4}\)
ⓒ. \(W\) and \(4W\)
ⓓ. \(\displaystyle \frac{3W}{2}\) and \(\displaystyle \frac{9W}{2}\)
Correct Answer: \(\displaystyle \frac{3W}{4}\) and \(\displaystyle \frac{15W}{4}\)
Explanation: The metre rule's own weight acts through the \(50\,\text{cm}\) mark.
\(\textbf{Moment balance about the pivot:}\)
The \(2W\) load is \(20\,\text{cm}\) to the left of the pivot:
\[
\tau_{\rm left}=2W(20)
\]
The rule's own weight acts \(10\,\text{cm}\) to the right:
\[
\tau_{\rm rule}=W(10)
\]
The unknown load acts \(40\,\text{cm}\) to the right:
\[
\tau_P=P(40)
\]
Therefore,
\[
2W(20)=W(10)+P(40)
\]
Thus,
\[
40W=10W+40P
\]
Hence,
\[
P=\frac{30W}{40}=\frac{3W}{4}
\]
\(\textbf{Vertical force balance:}\)
The pivot reaction \(R\) supports all downward forces:
\[
R=W+2W+\frac{3W}{4}
\]
Therefore,
\[
R=\frac{15W}{4}
\]
\(\textbf{Final answer:}\)
\[
\boxed{P=\frac{3W}{4},\qquad R=\frac{15W}{4}}
\]
413. A uniform rod of weight \(W\) and length \(L\) is hinged at its lower end and held at \(60^\circ\) above the horizontal. A horizontal force \(F\), directed to the left, is applied at its upper end so that the rod remains in equilibrium. The magnitude of \(F\) and the magnitude of the hinge reaction are
ⓐ. \(\displaystyle \frac{W}{2\sqrt3}\) and \(\displaystyle \frac{\sqrt{13}}{2\sqrt3}W\)
ⓑ. \(\displaystyle \frac{W}{\sqrt3}\) and \(\displaystyle \frac{2W}{\sqrt3}\)
ⓒ. \(\displaystyle \frac{\sqrt3W}{2}\) and \(\displaystyle \frac{\sqrt7}{2}W\)
ⓓ. \(\displaystyle \frac{W}{2}\) and \(\displaystyle \frac{\sqrt5}{2}W\)
Correct Answer: \(\displaystyle \frac{W}{2\sqrt3}\) and \(\displaystyle \frac{\sqrt{13}}{2\sqrt3}W\)
Explanation: Take moments about the hinge so that the hinge reaction does not enter the torque equation.
\(\textbf{Moment produced by the horizontal force:}\)
Its perpendicular moment arm is the vertical height of the upper end:
\[
L\sin60^\circ
\]
Hence its moment magnitude is
\[
FL\sin60^\circ
\]
The rod weight acts at its midpoint.
Its horizontal lever arm is
\[
\frac L2\cos60^\circ
\]
Thus rotational equilibrium requires
\[
FL\sin60^\circ
=W\frac L2\cos60^\circ
\]
Therefore,
\[
F=\frac W2\cot60^\circ
=\frac{W}{2\sqrt3}
\]
\(\textbf{Force balance at the hinge:}\)
The hinge must balance the horizontal force, so
\[
H_x=F=\frac{W}{2\sqrt3}
\]
Vertically,
\[
H_y=W
\]
Thus,
\[
H=\sqrt{H_x^2+H_y^2}
\]
Hence,
\[
H=W\sqrt{1+\frac1{12}}
=W\sqrt{\frac{13}{12}}
=\frac{\sqrt{13}}{2\sqrt3}W
\]
\(\textbf{Final answer:}\)
\[
\boxed{F=\frac{W}{2\sqrt3},\qquad H=\frac{\sqrt{13}}{2\sqrt3}W}
\]
414. A horizontal beam of length \(6\,\text{m}\) and weight \(100\,\text{N}\) is acted upon by vertical reactions at A and B, located \(1\,\text{m}\) and \(5\,\text{m}\) from the left end, respectively. A \(300\,\text{N}\) load hangs from the right end. If the supports are capable of exerting vertical forces in either direction, the reactions at A and B are
ⓐ. \(25\,\text{N}\) upward and \(375\,\text{N}\) upward
ⓑ. \(25\,\text{N}\) downward and \(425\,\text{N}\) upward
ⓒ. \(100\,\text{N}\) downward and \(500\,\text{N}\) upward
ⓓ. \(75\,\text{N}\) upward and \(325\,\text{N}\) upward
Correct Answer: \(25\,\text{N}\) downward and \(425\,\text{N}\) upward
Explanation: Let upward reactions be positive.
The beam's own weight acts at its centre, \(3\,\text{m}\) from the left end.
\(\textbf{Take moments about A:}\)
The distance \(AB\) is
\[
5-1=4\,\text{m}
\]
The beam weight is \(2\,\text{m}\) to the right of A, while the \(300\,\text{N}\) load is \(5\,\text{m}\) to the right.
Therefore,
\[
R_B(4)=100(2)+300(5)
\]
Hence,
\[
4R_B=200+1500=1700
\]
Thus,
\[
R_B=425\,\text{N}
\]
\(\textbf{Vertical force balance:}\)
\[
R_A+425-100-300=0
\]
Therefore,
\[
R_A=-25\,\text{N}
\]
The negative sign means the reaction at A must act downward.
If A were an ordinary contact support incapable of pulling downward, it would lose contact instead of supplying this reaction.
\(\textbf{Final answer:}\)
\[
R_A=25\,\text{N downward},\qquad R_B=425\,\text{N upward}
\]
415. For a coplanar force system acting on a rigid body, the net torque is found to be zero about two distinct points A and B. Which conclusion is necessarily correct?
ⓐ. The resultant force must be zero
ⓑ. A non-zero resultant force must have its line of action along AB
ⓒ. A non-zero resultant force must have its line of action perpendicular to AB
ⓓ. A non-zero resultant force may have any line of action in the plane
Correct Answer: A non-zero resultant force must have its line of action along AB
Explanation: Torque about different origins is related through the resultant force.
If the net torques about A and B are both zero, then the change in torque produced by shifting the reference point from A to B is also zero.
Therefore,
\[
\vec r_{AB}\times\vec F_{\rm resultant}=0.
\]
This condition allows two possibilities.
The resultant force may itself be zero.
Alternatively, a non-zero resultant force may be parallel to \(\vec r_{AB}\).
Since its torque is zero about A as well, its line of action must pass through A.
A line through A parallel to AB is the line AB itself.
Thus a non-zero resultant may act along AB while producing zero torque about both A and B.
Consequently, zero torque about two points does not by itself prove complete equilibrium.
416. A horizontal rod of weight \(W\) is suspended by two vertical strings attached at points \(L/4\) and \(3L/4\) from its left end. A load \(2W\) is attached at the right end of the rod. Which conclusion follows from the equilibrium equations if both strings are assumed to remain taut?
ⓐ. \(T_L=W/2,\quad T_R=5W/2\), so both strings can remain taut
ⓑ. \(T_L=W,\quad T_R=2W\), so both strings can remain taut
ⓒ. \(T_L=-W/2,\quad T_R=7W/2\), so equilibrium with both strings taut is impossible
ⓓ. \(T_L=7W/2,\quad T_R=-W/2\), so only the right string cannot remain taut
Correct Answer: \(T_L=-W/2,\quad T_R=7W/2\), so equilibrium with both strings taut is impossible
Explanation: Let the left and right string tensions be \(T_1\) and \(T_2\).
The rod's own weight acts at its midpoint.
\(\textbf{Moment balance about the left string attachment:}\)
The right string is a distance \(L/2\) away:
\[
T_2\left(\frac L2\right).
\]
The rod's weight acts \(L/4\) to the right:
\[
W\left(\frac L4\right).
\]
The \(2W\) load is \(3L/4\) to the right:
\[
2W\left(\frac{3L}{4}\right).
\]
Thus,
\[
T_2\frac L2
=W\frac L4+2W\frac{3L}{4}.
\]
Hence,
\[
\frac{T_2}{2}
=\frac W4+\frac{3W}{2}
=\frac{7W}{4}.
\]
Therefore,
\[
T_2=\frac{7W}{2}.
\]
\(\textbf{Vertical force balance:}\)
\[
T_1+T_2=3W.
\]
Thus,
\[
T_1=3W-\frac{7W}{2}
=-\frac W2.
\]
A string cannot exert a downward push, so a negative tension is physically impossible.
The assumed two-taut-string equilibrium therefore cannot occur.
\(\textbf{Final answer:}\) The left string would have to supply \(-W/2\), so the assumed equilibrium is impossible.
417. A uniform horizontal rod of weight \(W\) is hinged at its left end and supported at its right end by a string making an unknown angle \(\theta\) above the rod. In equilibrium, the horizontal component of the hinge reaction has magnitude \(\sqrt3W/2\). The angle \(\theta\) and the string tension are
ⓐ. \(30^\circ\) and \(W\)
ⓑ. \(60^\circ\) and \(W\)
ⓒ. \(30^\circ\) and \(2W\)
ⓓ. \(45^\circ\) and \(W/\sqrt2\)
Correct Answer: \(30^\circ\) and \(W\)
Explanation: Take moments about the hinge.
The rod's weight acts at its midpoint.
\(\textbf{Torque balance:}\)
Only the vertical component of the string tension produces a moment about the hinge:
\[
T\sin\theta\,L
=W\left(\frac L2\right)
\]
Therefore,
\[
T\sin\theta=\frac W2. \tag{1}
\]
\(\textbf{Horizontal force balance:}\)
The string's horizontal component is balanced by the hinge reaction:
\[
T\cos\theta=\frac{\sqrt3W}{2}. \tag{2}
\]
Dividing Eq. (2) by Eq. (1),
\[
\cot\theta=\sqrt3
\]
Hence,
\[
\theta=30^\circ
\]
Using Eq. (1),
\[
T\sin30^\circ=\frac W2
\]
Therefore,
\[
\frac T2=\frac W2,
\]
so
\[
T=W
\]
\(\textbf{Final answer:}\)
\[
\boxed{\theta=30^\circ,\qquad T=W}
\]
418. Consider the following statements for planar rigid-body equilibrium.
Statement I: If \(\sum\vec F=0\), verifying \(\sum\tau=0\) about any one chosen point is sufficient for rotational equilibrium.
Statement II: When \(\sum\vec F=0\), the net torque is independent of the chosen origin.
Statement III: A support force passing through the chosen moment point may be omitted from that moment equation, but it must still be included in force balance.
Statement IV: If \(\sum\tau=0\) about one point, every individual external force must have zero moment about that point.
Which combination is correct?
ⓐ. I and II only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Statement I is correct because once the resultant external force is zero, torque balance about one point establishes the rotational equilibrium condition.
Statement II is also correct.
When
\[
\sum\vec F=0,
\]
changing the origin does not change the net torque.
Statement III is an important calculation rule.
A force whose line of action passes through the chosen moment point contributes zero torque there.
However, the force still physically acts and must be retained when writing
\[
\sum\vec F=0.
\]
Statement IV is false.
Several non-zero individual torques can cancel so that their algebraic sum is zero.
Rotational equilibrium requires the net torque to vanish, not every separate torque.
Therefore Statements I, II and III are correct.
419. A rigid body is acted upon by exactly two external forces and is in complete mechanical equilibrium. Which statement must be true?
ⓐ. The forces must have equal magnitudes, with no restriction on their directions
ⓑ. The forces must be equal in magnitude, opposite in direction and collinear
ⓒ. The forces must be equal in magnitude and opposite, but may be non-collinear
ⓓ. The forces may have unequal magnitudes if they are opposite and collinear
Correct Answer: The forces must be equal in magnitude, opposite in direction and collinear
Explanation: Complete equilibrium first requires zero resultant force.
With only two external forces, this means the forces must have equal magnitudes and opposite directions.
However, this condition alone is not sufficient.
If their lines of action are parallel but distinct, the two forces form a couple.
Such a couple has zero resultant force but a non-zero net torque.
Rotational equilibrium therefore additionally requires zero moment.
For two equal and opposite forces, that is possible only when their lines of action coincide.
Hence the two forces must be collinear as well as equal and opposite.
This is the characteristic equilibrium condition for a two-force rigid body.
A separated equal-and-opposite pair would violate rotational equilibrium.
420. A rigid rectangular plate lies in the \(xy\)-plane. Forces act at the following points:
\[
\vec r_A=(0,0),\qquad \vec F_A=6\hat i\,\text{N},
\]
\[
\vec r_B=2\hat j\,\text{m},\qquad \vec F_B=-2\hat i\,\text{N},
\]
\[
\vec r_C=4\hat i\,\text{m},\qquad \vec F_C=(-4\hat i+3\hat j)\,\text{N}.
\]
An additional force \(\vec F_D\) acts at
\[
\vec r_D=(4\hat i+2\hat j)\,\text{m},
\]
and an external couple is also applied. For complete equilibrium, \(\vec F_D\) and the required couple are
ⓐ. \(-3\hat j\,\text{N}\) and \(4\,\text{N m}\) clockwise
ⓑ. \(-3\hat j\,\text{N}\) and \(4\,\text{N m}\) anticlockwise
ⓒ. \(+3\hat j\,\text{N}\) and \(4\,\text{N m}\) clockwise
ⓓ. \(-3\hat j\,\text{N}\) and \(8\,\text{N m}\) clockwise
Correct Answer: \(-3\hat j\,\text{N}\) and \(4\,\text{N m}\) clockwise
Explanation: Complete equilibrium requires both zero resultant force and zero resultant torque.
\(\textbf{First enforce force balance:}\)
The sum of the three known forces is
\[
\vec F_A+\vec F_B+\vec F_C
=6\hat i-2\hat i-4\hat i+3\hat j
\]
Therefore,
\[
\vec F_{\rm known}=3\hat j\,\text{N}
\]
Hence the fourth force must be
\[
\vec F_D=-3\hat j\,\text{N}
\]
\(\textbf{Torque of the known and balancing forces about A:}\)
Force A has zero moment because it acts at A.
For B,
\[
\tau_{Bz}=xF_y-yF_x=0-(2)(-2)=+4\,\text{N m}
\]
For C,
\[
\tau_{Cz}=(4)(3)-0=+12\,\text{N m}
\]
For D,
\[
\tau_{Dz}=(4)(-3)-(2)(0)=-12\,\text{N m}
\]
Thus,
\[
\tau_{\rm forces}=4+12-12=+4\,\text{N m}
\]
The positive sign corresponds to an anticlockwise moment.
The external couple must therefore supply
\[
-4\,\text{N m},
\]
that is, \(4\,\text{N m}\) clockwise.
\(\textbf{Final answer:}\)
\[
\vec F_D=-3\hat j\,\text{N},
\]
with a \(4\,\text{N m}\) clockwise couple.
