301. A body initially at rest separates into three fragments. A \(2\,\text{kg}\) fragment moves eastward at \(5\,\text{m s}^{-1}\). A \(1\,\text{kg}\) fragment moves at \(10\,\text{m s}^{-1}\), \(60^\circ\) north of west. The third fragment has mass \(5\,\text{kg}\). External impulse is negligible. Which option gives the third fragment's velocity?
ⓐ. \(\sqrt3\,\text{m s}^{-1}\), due south
ⓑ. \(2\,\text{m s}^{-1}\), \(60^\circ\) north of east
ⓒ. \(4\,\text{m s}^{-1}\), due west
ⓓ. \(2\,\text{m s}^{-1}\), \(60^\circ\) south of west
Correct Answer: \(2\,\text{m s}^{-1}\), \(60^\circ\) south of west
Explanation: Let east be \(+\hat i\) and north be \(+\hat j\). The first fragment has momentum
\[
\vec p_1=2(5)\hat i=10\hat i.
\]
The second fragment's momentum magnitude is
\[
p_2=1(10)=10\,\text{kg m s}^{-1}.
\]
At \(60^\circ\) north of west, its components are
\[
\vec p_2=-10\cos60^\circ\,\hat i+10\sin60^\circ\,\hat j.
\]
Therefore,
\[
\vec p_2=-5\hat i+5\sqrt3\hat j.
\]
The resultant of the first two momenta is
\[
\vec p_1+\vec p_2=5\hat i+5\sqrt3\hat j.
\]
Since the parent was initially at rest,
\[
\vec p_3=-5\hat i-5\sqrt3\hat j.
\]
For \(m_3=5\,\text{kg}\),
\[
\vec v_3=-\hat i-\sqrt3\hat j\,\text{m s}^{-1}.
\]
Its speed is
\[
v_3=\sqrt{1+3}=2\,\text{m s}^{-1}.
\]
Its components correspond to \(60^\circ\) south of west.
302. A \(9\,\text{kg}\) body is initially at rest on a smooth horizontal plane. In Stage I, it separates into a \(3\,\text{kg}\) fragment \(A\) moving westward at \(4\,\text{m s}^{-1}\) and a \(6\,\text{kg}\) remainder \(BC\). In Stage II, \(BC\) separates into a \(2\,\text{kg}\) fragment \(B\) moving northward at \(3\,\text{m s}^{-1}\) and a \(4\,\text{kg}\) fragment \(C\). External horizontal impulse is negligible throughout. What is the final velocity of \(C\)?
ⓐ. \((3\hat i-1.5\hat j)\,\text{m s}^{-1}\), speed \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
ⓑ. \((3\hat i+1.5\hat j)\,\text{m s}^{-1}\), speed \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
ⓒ. \((1.5\hat i-3\hat j)\,\text{m s}^{-1}\), speed \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
ⓓ. \((3\hat i-\hat j)\,\text{m s}^{-1}\), speed \(\sqrt{10}\,\text{m s}^{-1}\)
Correct Answer: \((3\hat i-1.5\hat j)\,\text{m s}^{-1}\), speed \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
Explanation: Take east as \(+\hat i\) and north as \(+\hat j\). The original \(9\,\text{kg}\) body is at rest, so its momentum is zero. In Stage I, fragment \(A\) has
\[
\vec p_A=3(-4\hat i)=-12\hat i.
\]
Therefore the remainder \(BC\) must carry
\[
\vec p_{BC}=+12\hat i.
\]
This momentum is the initial momentum of the Stage-II \(BC\) subsystem. Fragment \(B\) emerges with
\[
\vec p_B=2(3\hat j)=6\hat j.
\]
Hence,
\[
\vec p_C=\vec p_{BC}-\vec p_B
=12\hat i-6\hat j.
\]
Since \(m_C=4\,\text{kg}\),
\[
\vec v_C=3\hat i-1.5\hat j\,\text{m s}^{-1}.
\]
Its speed is
\[
v_C=\sqrt{3^2+1.5^2}
=\sqrt{\frac{45}{4}}
=\frac{3\sqrt5}{2}\,\text{m s}^{-1}.
\]
The momentum of the intermediate \(BC\) body must be preserved into Stage II rather than assuming the second separation starts from rest.
303. A \(10\,\text{kg}\) body initially has velocity
\[
\vec u=(2\hat i+\hat j)\,\text{m s}^{-1}.
\]
During a short fragmentation event, the complete system receives an external impulse
\[
\vec J_{\text{ext}}=(-4\hat i+2\hat j)\,\text{N s}.
\]
Which row gives a set of final fragment momenta compatible with the system momentum balance?
| Case | \(\vec p_1\) | \(\vec p_2\) | \(\vec p_3\) |
|---|
| P | \(6\hat i+2\hat j\) | \(4\hat i+5\hat j\) | \(6\hat i+5\hat j\) |
| Q | \(6\hat i+2\hat j\) | \(4\hat i+5\hat j\) | \(8\hat i+5\hat j\) |
| R | \(6\hat i+2\hat j\) | \(4\hat i+3\hat j\) | \(6\hat i+5\hat j\) |
| S | \(4\hat i+2\hat j\) | \(4\hat i+5\hat j\) | \(6\hat i+5\hat j\) |
ⓐ. R
ⓑ. Q
ⓒ. P
ⓓ. S
Correct Answer: P
Explanation: The initial momentum of the parent body is
\[
\vec P_i=10(2\hat i+\hat j)
=20\hat i+10\hat j.
\]
The external impulse changes the complete system momentum to
\[
\vec P_f=\vec P_i+\vec J_{\text{ext}}.
\]
Thus,
\[
\vec P_f
=(20\hat i+10\hat j)+(-4\hat i+2\hat j)
=16\hat i+12\hat j.
\]
For case P,
\[
\vec p_1+\vec p_2+\vec p_3
=(6+4+6)\hat i+(2+5+5)\hat j.
\]
Hence,
\[
\vec P_P=16\hat i+12\hat j,
\]
which matches the required final total. Case Q gives \(18\hat i+12\hat j\). Case R gives \(16\hat i+10\hat j\). Case S gives \(14\hat i+12\hat j\). Therefore only case P is consistent with the impulse-momentum balance of the complete fragmented system.
304. Consider the following statements about recoil, push-off and fragmentation processes.
Statement I: If an initially stationary isolated body separates into three fragments, the momentum of each fragment need not have a single equal-and-opposite partner; only the vector sum of all fragment momenta must be zero.
Statement II: If a system receives zero external impulse in the \(x\)-direction but a non-zero external impulse in the \(y\)-direction, its total \(x\)-momentum may be conserved even though its complete momentum vector is not.
Statement III: Internal forces can alter the momentum of each part of a system while leaving the complete system momentum unchanged when the net external impulse is zero.
Statement IV: In a recoil problem, equal and opposite final velocities are required whenever the initial total momentum is zero.
Which option is correct?
ⓐ. II and III only
ⓑ. I and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Statement I is correct because momentum conservation constrains the total vector sum:
\[
\vec p_1+\vec p_2+\vec p_3=0
\]
for an initially stationary isolated three-fragment system. The fragments do not have to form separate one-to-one equal-and-opposite pairs. Statement II is also correct because momentum conservation can be applied component-wise. If
\[
J_{\text{ext},x}=0,
\]
then total \(P_x\) remains unchanged even when a non-zero \(J_{\text{ext},y}\) changes \(P_y\). Statement III is correct because internal impulses redistribute momentum among the parts while cancelling in the complete-system total. Statement IV is incorrect. Zero initial total momentum requires equal and opposite final momenta in a two-body recoil process:
\[
M\vec V=-m\vec v.
\]
If \(M\neq m\), the velocity magnitudes are not equal. Thus equal-and-opposite velocities are a special equal-mass case, not a general consequence of momentum conservation.
305. Which statement correctly describes a set of concurrent forces?
ⓐ. Their lines of action pass through a common point.
ⓑ. They must all have the same magnitude.
ⓒ. They must all act in mutually perpendicular directions.
ⓓ. Any forces acting at different points of a rigid body are necessarily concurrent.
Correct Answer: Their lines of action pass through a common point.
Explanation: Concurrent forces are identified by the geometry of their lines of action. For a concurrent-force system, those lines pass through one common point. The forces need not have equal magnitudes. They also need not be perpendicular to one another. Two, three or more concurrent forces may act at many different angles. Their resultant is obtained by vector addition:
\[
\vec R=\sum\vec F.
\]
Equal magnitudes alone do not make forces concurrent and do not guarantee zero resultant. Forces acting at separated points on an extended rigid body are not automatically concurrent because their lines of action may fail to meet at one common point. In the particle model used for translational equilibrium, forces are conveniently represented as acting through the particle. Hence option A gives the defining condition.
306. Two concurrent forces of \(6\,\text{N}\) eastward and \(8\,\text{N}\) northward act on a particle. What is their resultant?
ⓐ. \(14\,\text{N}\), \(45^\circ\) north of east
ⓑ. \(10\,\text{N}\), approximately \(53^\circ\) north of east
ⓒ. \(10\,\text{N}\), approximately \(37^\circ\) north of east
ⓓ. \(2\,\text{N}\), northward
Correct Answer: \(10\,\text{N}\), approximately \(53^\circ\) north of east
Explanation: The two forces are perpendicular, so they form the rectangular components of the resultant:
\[
R_x=6\,\text{N},\qquad R_y=8\,\text{N}.
\]
The magnitude is
\[
R=\sqrt{R_x^2+R_y^2}.
\]
Thus,
\[
R=\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{N}.
\]
If \(\theta\) is measured north of east,
\[
\tan\theta=\frac{R_y}{R_x}=\frac{8}{6}=\frac43.
\]
Hence,
\[
\theta\approx53^\circ.
\]
Adding the force magnitudes arithmetically to obtain \(14\,\text{N}\) would ignore their different directions. Therefore the resultant is \(10\,\text{N}\), approximately \(53^\circ\) north of east.
307. Two concurrent forces, each of magnitude \(10\,\text{N}\), act with an angle of \(120^\circ\) between them. What is the magnitude and direction of their resultant?
ⓐ. Zero
ⓑ. \(20\,\text{N}\), along either force
ⓒ. \(10\sqrt3\,\text{N}\), perpendicular to the bisector
ⓓ. \(10\,\text{N}\), along the internal angle bisector
Correct Answer: \(10\,\text{N}\), along the internal angle bisector
Explanation: For two forces \(F_1\) and \(F_2\) separated by angle \(\theta\),
\[
R^2=F_1^2+F_2^2+2F_1F_2\cos\theta.
\]
Here,
\[
F_1=F_2=10\,\text{N},\qquad \theta=120^\circ.
\]
Therefore,
\[
R^2=100+100+200\cos120^\circ.
\]
Since
\[
\cos120^\circ=-\frac12,
\]
we obtain
\[
R^2=200-100=100.
\]
Thus,
\[
R=10\,\text{N}.
\]
Because the two original forces have equal magnitudes, their resultant lies along their internal angular bisector. The forces do not cancel because they are not opposite; cancellation would require an angle of \(180^\circ\).
308. Four concurrent forces act on a particle: \(12\,\text{N}\) eastward, \(5\,\text{N}\) northward, \(7\,\text{N}\) westward and \(1\,\text{N}\) southward. What is the resultant?
ⓐ. \(5\,\text{N}\), eastward
ⓑ. \(9\,\text{N}\), north-east
ⓒ. \(\sqrt{41}\,\text{N}\), at \(\tan^{-1}(4/5)\) north of east
ⓓ. \(\sqrt{41}\,\text{N}\), at \(\tan^{-1}(5/4)\) north of east
Correct Answer: \(\sqrt{41}\,\text{N}\), at \(\tan^{-1}(4/5)\) north of east
Explanation: Take east and north as the positive \(x\)- and \(y\)-directions. The horizontal resultant component is
\[
R_x=12-7=5\,\text{N}.
\]
The vertical component is
\[
R_y=5-1=4\,\text{N}.
\]
Therefore the resultant magnitude is
\[
R=\sqrt{5^2+4^2}=\sqrt{41}\,\text{N}.
\]
Both components are positive, so the resultant lies in the north-east quadrant. If \(\theta\) is measured north of east,
\[
\tan\theta=\frac{R_y}{R_x}=\frac45.
\]
Hence,
\[
\theta=\tan^{-1}\left(\frac45\right).
\]
The individual force magnitudes should not be directly added because opposite components partially cancel. Component addition preserves both magnitude and directional information.
309. Three concurrent forces act on a particle as shown in the table. Take \(\cos53^\circ=0.6\) and \(\sin53^\circ=0.8\).
| Force | Magnitude | Direction |
|---|
| \(F_1\) | \(10\,\text{N}\) | East |
| \(F_2\) | \(4\,\text{N}\) | North |
| \(F_3\) | \(10\,\text{N}\) | \(53^\circ\) south of west |
What is the resultant?
ⓐ. \(8\,\text{N}\), westward
ⓑ. \(4\sqrt2\,\text{N}\), \(45^\circ\) north of east
ⓒ. \(4\sqrt2\,\text{N}\), \(45^\circ\) south of east
ⓓ. \(10\,\text{N}\), south-west
Correct Answer: \(4\sqrt2\,\text{N}\), \(45^\circ\) south of east
Explanation: Resolve \(F_3\) into rectangular components. Since it is \(53^\circ\) south of west,
\[
F_{3x}=-10\cos53^\circ=-6\,\text{N},
\]
and
\[
F_{3y}=-10\sin53^\circ=-8\,\text{N}.
\]
The total horizontal component is
\[
R_x=10-6=4\,\text{N}.
\]
The total vertical component is
\[
R_y=4-8=-4\,\text{N}.
\]
Therefore,
\[
R=\sqrt{4^2+(-4)^2}=4\sqrt2\,\text{N}.
\]
A positive \(x\)-component and negative \(y\)-component place the resultant in the south-east quadrant. Since the component magnitudes are equal, the angle is \(45^\circ\). Hence the resultant is \(4\sqrt2\,\text{N}\), \(45^\circ\) south of east.
310. Two concurrent forces are
\[
\vec F_1=(4\hat i+3\hat j)\,\text{N}
\]
and
\[
\vec F_2=(-2\hat i+5\hat j)\,\text{N}.
\]
A third force \(\vec F_3\) is added so that the final resultant becomes
\[
\vec R=(5\hat i+4\hat j)\,\text{N}.
\]
What is \(\vec F_3\)?
ⓐ. \((3\hat i-4\hat j)\,\text{N}\), magnitude \(5\,\text{N}\)
ⓑ. \((3\hat i+4\hat j)\,\text{N}\), magnitude \(5\,\text{N}\)
ⓒ. \((-3\hat i+4\hat j)\,\text{N}\), magnitude \(5\,\text{N}\)
ⓓ. \((7\hat i-4\hat j)\,\text{N}\), magnitude \(\sqrt{65}\,\text{N}\)
Correct Answer: \((3\hat i-4\hat j)\,\text{N}\), magnitude \(5\,\text{N}\)
Explanation: The three forces must satisfy
\[
\vec F_1+\vec F_2+\vec F_3=\vec R.
\]
First add the two known forces:
\[
\vec F_1+\vec F_2=(4-2)\hat i+(3+5)\hat j.
\]
Thus,
\[
\vec F_1+\vec F_2=2\hat i+8\hat j.
\]
Therefore,
\[
\vec F_3=\vec R-(\vec F_1+\vec F_2).
\]
Substituting,
\[
\vec F_3=(5\hat i+4\hat j)-(2\hat i+8\hat j)
=3\hat i-4\hat j.
\]
Its magnitude is
\[
F_3=\sqrt{3^2+(-4)^2}=5\,\text{N}.
\]
The force points south of east because its \(x\)-component is positive while its \(y\)-component is negative.
311. Two concurrent forces are \(5\,\text{N}\) eastward and \(12\,\text{N}\) northward. A third force is added so that the three force vectors form a closed triangle and have zero resultant. If this third force is then reversed without changing its magnitude, what is the new resultant?
ⓐ. \(13\,\text{N}\), north-east
ⓑ. \(26\,\text{N}\) north-east
ⓒ. \(13\,\text{N}\), south-west
ⓓ. \(24\,\text{N}\), northward
Correct Answer: \(26\,\text{N}\) north-east
Explanation: The resultant of the first two forces is
\[
\vec R_{12}=5\hat i+12\hat j.
\]
Its magnitude is
\[
R_{12}=\sqrt{5^2+12^2}=13\,\text{N}.
\]
For the original three forces to form a closed force triangle,
\[
\vec F_3=-\vec R_{12}.
\]
Thus the third force is \(13\,\text{N}\) in the opposite direction to \(\vec R_{12}\). Reversing it gives
\[
\vec F_3'=\vec R_{12}.
\]
The new resultant is therefore
\[
\vec R'=\vec R_{12}+\vec F_3'=2\vec R_{12}.
\]
Hence,
\[
R'=2(13)=26\,\text{N}.
\]
Its direction is the same as that of the original resultant of the \(5\,\text{N}\) and \(12\,\text{N}\) forces. Reversing a balancing vector therefore doubles rather than preserves the original pair resultant.
312. In arrangement \(P\), two equal concurrent forces of magnitude \(F\) have an angle of \(60^\circ\) between them. In arrangement \(Q\), the same two force magnitudes have an angle of \(120^\circ\) between them. What is \(R_P:R_Q\)?
ⓐ. \(1:\sqrt3\)
ⓑ. \(\sqrt3:1\)
ⓒ. \(2:1\)
ⓓ. \(1:1\)
Correct Answer: \(\sqrt3:1\)
Explanation: For two equal forces \(F\) separated by angle \(\theta\),
\[
R=\sqrt{F^2+F^2+2F^2\cos\theta}.
\]
For arrangement \(P\),
\[
R_P=F\sqrt{2+2\cos60^\circ}.
\]
Since \(\cos60^\circ=1/2\),
\[
R_P=F\sqrt3.
\]
For arrangement \(Q\),
\[
R_Q=F\sqrt{2+2\cos120^\circ}.
\]
Using \(\cos120^\circ=-1/2\),
\[
R_Q=F.
\]
Therefore,
\[
R_P:R_Q=\sqrt3:1.
\]
Increasing the angle between fixed-magnitude forces reduces the magnitude of their resultant because their components oppose each other more strongly.
313. Consider the following statements about concurrent forces.
Statement I: If \(\sum F_x=0\) but \(\sum F_y\neq0\), the resultant is not zero and lies along the \(y\)-direction.
Statement II: Two concurrent forces of equal magnitude always have zero resultant.
Which option is correct?
ⓐ. Both Statement I and Statement II are correct.
ⓑ. Both Statement I and Statement II are incorrect.
ⓒ. Statement I is incorrect but Statement II is correct.
ⓓ. Statement I is correct but Statement II is incorrect.
Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: The resultant of concurrent forces can be reconstructed from its components:
\[
\vec R=R_x\hat i+R_y\hat j.
\]
If
\[
R_x=\sum F_x=0
\]
but
\[
R_y=\sum F_y\neq0,
\]
then
\[
\vec R=R_y\hat j,
\]
so the resultant is non-zero and lies along the \(y\)-axis. Statement I is therefore correct. Equal magnitudes alone do not make two forces cancel. Two forces have zero resultant only when they are equal in magnitude and opposite in direction, corresponding to an angle of \(180^\circ\) between them. If equal forces are perpendicular, for example, their resultant is \(\sqrt2F\), not zero. Hence Statement II is incorrect.
314. Match the concurrent-force arrangement in List I with the magnitude of its resultant in List II. Each force mentioned has magnitude \(F\).
| List I | Arrangement | List II | Resultant magnitude |
|---|
| P | Two forces in the same direction | 1 | \(0\) |
| Q | Two forces in opposite directions | 2 | \(F\) |
| R | Two forces at \(90^\circ\) | 3 | \(\sqrt2F\) |
| S | Two forces at \(120^\circ\) | 4 | \(2F\) |
Choose the correct matching.
ⓐ. P-4, Q-1, R-3, S-2
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-3, Q-1, R-2, S-4
Correct Answer: P-4, Q-1, R-3, S-2
Explanation: For two forces in the same direction, their vectors add directly:
\[
R=F+F=2F,
\]
so \(P\rightarrow4\). For opposite directions, equal forces cancel:
\[
R=F-F=0,
\]
so \(Q\rightarrow1\). For perpendicular forces,
\[
R=\sqrt{F^2+F^2}=\sqrt2F,
\]
hence \(R\rightarrow3\). At \(120^\circ\),
\[
R^2=F^2+F^2+2F^2\cos120^\circ.
\]
Thus,
\[
R^2=2F^2-F^2=F^2,
\]
so \(R=F\) and \(S\rightarrow2\). The matching is therefore P-4, Q-1, R-3, S-2.
315. Two equal concurrent forces, each of magnitude \(F\), have a resultant of magnitude \(F\). One of the forces is then doubled in magnitude without changing either force direction. What is the new resultant magnitude?
ⓐ. \(F\)
ⓑ. \(\sqrt2F\)
ⓒ. \(\sqrt3F\)
ⓓ. \(2F\)
Correct Answer: \(\sqrt3F\)
Explanation: First use the original condition to determine the angle \(\theta\) between the forces. For two equal forces,
\[
R^2=F^2+F^2+2F^2\cos\theta.
\]
Given \(R=F\),
\[
F^2=2F^2+2F^2\cos\theta.
\]
Dividing by \(F^2\),
\[
1=2+2\cos\theta.
\]
Therefore,
\[
\cos\theta=-\frac12,
\]
so
\[
\theta=120^\circ.
\]
After one force becomes \(2F\), the new resultant satisfies
\[
R'^2=F^2+(2F)^2+2(F)(2F)\cos120^\circ.
\]
Hence,
\[
R'^2=F^2+4F^2-2F^2=3F^2.
\]
Thus,
\[
R'=\sqrt3F.
\]
316. Three concurrent forces of equal magnitude \(F\) act at angular separations of \(120^\circ\) from one another, so their initial resultant is zero. The magnitude of one force is then doubled while its direction and the other two forces remain unchanged. What is the new resultant?
ⓐ. Zero
ⓑ. Magnitude \(F\), along the direction of the force that was doubled
ⓒ. Magnitude \(2F\), opposite to the doubled force
ⓓ. Magnitude \(\sqrt3F\), perpendicular to the doubled force
Correct Answer: Magnitude \(F\), along the direction of the force that was doubled
Explanation: Let the original forces be \(\vec F_1,\vec F_2,\vec F_3\). Since the three equal forces are \(120^\circ\) apart,
\[
\vec F_1+\vec F_2+\vec F_3=0.
\]
Suppose \(\vec F_1\) is doubled. The new resultant becomes
\[
\vec R'=2\vec F_1+\vec F_2+\vec F_3.
\]
Use the original zero-resultant relation:
\[
\vec F_2+\vec F_3=-\vec F_1.
\]
Therefore,
\[
\vec R'=2\vec F_1-\vec F_1=\vec F_1.
\]
Hence the new resultant has magnitude
\[
R'=F
\]
and points in the original direction of the force whose magnitude was doubled. The change from \(F\) to \(2F\) effectively adds one extra copy of that vector to a set whose original sum was zero.
317. Two concurrent forces have magnitudes \(10\,\text{N}\) and \(14\,\text{N}\). Their resultant has magnitude \(20\,\text{N}\). The \(14\,\text{N}\) force is now reversed while the \(10\,\text{N}\) force remains unchanged. What is the magnitude of the new resultant?
ⓐ. \(4\,\text{N}\)
ⓑ. \(8\sqrt3\,\text{N}\)
ⓒ. \(12\sqrt2\,\text{N}\)
ⓓ. \(24\,\text{N}\)
Correct Answer: \(8\sqrt3\,\text{N}\)
Explanation: Let the original angle between the two forces be \(\theta\). The original resultant satisfies
\[
20^2=10^2+14^2+2(10)(14)\cos\theta.
\]
Thus,
\[
400=100+196+280\cos\theta.
\]
Therefore,
\[
280\cos\theta=104,
\]
so
\[
\cos\theta=\frac{13}{35}.
\]
Reversing the \(14\,\text{N}\) force changes its direction by \(180^\circ\). The cosine of the new angle is therefore
\[
-\cos\theta=-\frac{13}{35}.
\]
Hence,
\[
R'^2=10^2+14^2-280\left(\frac{13}{35}\right).
\]
So,
\[
R'^2=296-104=192.
\]
Therefore,
\[
R'=\sqrt{192}=8\sqrt3\,\text{N}.
\]
The changed vector direction, not merely the unchanged magnitudes, controls the new resultant.
318. Three concurrent forces are
\[
\vec F_1=(6\hat i+8\hat j)\,\text{N},
\qquad
\vec F_2=(-10\hat i+2\hat j)\,\text{N},
\]
and
\[
\vec F_3=(\lambda\hat i-5\hat j)\,\text{N}.
\]
The resultant is known to point due north. What are \(\lambda\) and the magnitude of the resultant?
ⓐ. \(\lambda=4,\ R=15\,\text{N}\)
ⓑ. \(\lambda=-4,\ R=5\,\text{N}\)
ⓒ. \(\lambda=4,\ R=5\,\text{N}\)
ⓓ. \(\lambda=10,\ R=5\,\text{N}\)
Correct Answer: \(\lambda=4,\ R=5\,\text{N}\)
Explanation: For the resultant to point due north, its horizontal component must be zero. The total \(x\)-component is
\[
R_x=6-10+\lambda.
\]
Hence,
\[
-4+\lambda=0,
\]
so
\[
\lambda=4.
\]
Now add the vertical components:
\[
R_y=8+2-5=5\,\text{N}.
\]
Because
\[
R_x=0
\]
and
\[
R_y=+5\,\text{N},
\]
the resultant is indeed directed due north. Its magnitude is therefore
\[
R=5\,\text{N}.
\]
The directional information provided in the stem acts as a component constraint that must be used before the resultant magnitude is determined.
319. Three concurrent forces have a resultant of magnitude \(10\,\text{N}\) directed \(37^\circ\) north of east. Two of the forces are \(3\,\text{N}\) eastward and \(4\,\text{N}\) northward. Take \(\cos37^\circ=0.8\) and \(\sin37^\circ=0.6\). What is the third force?
ⓐ. \((2\hat i+5\hat j)\,\text{N}\), magnitude \(\sqrt{29}\,\text{N}\)
ⓑ. \((5\hat i-2\hat j)\,\text{N}\), magnitude \(\sqrt{29}\,\text{N}\)
ⓒ. \((8\hat i+6\hat j)\,\text{N}\), magnitude \(10\,\text{N}\)
ⓓ. \((5\hat i+2\hat j)\,\text{N}\), magnitude \(\sqrt{29}\,\text{N}\)
Correct Answer: \((5\hat i+2\hat j)\,\text{N}\), magnitude \(\sqrt{29}\,\text{N}\)
Explanation: Convert the given resultant into rectangular components. Its eastward component is
\[
R_x=10\cos37^\circ=8\,\text{N}.
\]
Its northward component is
\[
R_y=10\sin37^\circ=6\,\text{N}.
\]
Thus,
\[
\vec R=8\hat i+6\hat j.
\]
The two known forces sum to
\[
\vec F_1+\vec F_2=3\hat i+4\hat j.
\]
Hence the third force is
\[
\vec F_3=\vec R-(\vec F_1+\vec F_2).
\]
Therefore,
\[
\vec F_3=(8-3)\hat i+(6-4)\hat j
=5\hat i+2\hat j.
\]
Its magnitude is
\[
F_3=\sqrt{5^2+2^2}=\sqrt{29}\,\text{N}.
\]
The problem requires converting a resultant representation into components before performing vector subtraction.
320. Three concurrent forces initially have zero resultant. Two of them are \(30\,\text{N}\) eastward and \(40\,\text{N}\) northward. The third force is then rotated through \(90^\circ\) clockwise without changing its magnitude, while the first two forces remain unchanged. What is the new resultant?
At the particle, the first force points horizontally east with magnitude \(30\,\text{N}\), the second points vertically north with magnitude \(40\,\text{N}\), and the original third force is the single vector required to balance these two. Only this third vector is subsequently rotated \(90^\circ\) clockwise.
ⓐ. \(50\,\text{N}\), directed \(90^\circ\) north of east
ⓑ. \(50\sqrt2\,\text{N}\), directed \(\tan^{-1}7\) north of west
ⓒ. \(70\,\text{N}\), directed due north
ⓓ. \(10\sqrt2\,\text{N}\), directed \(45^\circ\) north of west
Correct Answer: \(50\sqrt2\,\text{N}\), directed \(\tan^{-1}7\) north of west
Explanation: The resultant of the first two forces is
\[
\vec R_{12}=30\hat i+40\hat j.
\]
Its magnitude is
\[
R_{12}=\sqrt{30^2+40^2}=50\,\text{N}.
\]
For the original resultant of all three forces to be zero, the third force must be
\[
\vec F_3=-30\hat i-40\hat j.
\]
A \(90^\circ\) clockwise rotation transforms a vector \((x,y)\) into \((y,-x)\). Therefore the rotated third force is
\[
\vec F_3'=-40\hat i+30\hat j.
\]
The new resultant is
\[
\vec R'=(30\hat i+40\hat j)+(-40\hat i+30\hat j).
\]
Hence,
\[
\vec R'=-10\hat i+70\hat j.
\]
Its magnitude is
\[
R'=\sqrt{(-10)^2+70^2}=\sqrt{5000}=50\sqrt2\,\text{N}.
\]
The vector lies north-west, and if \(\theta\) is measured north of west,
\[
\tan\theta=\frac{70}{10}=7.
\]
Thus the new resultant is \(50\sqrt2\,\text{N}\), at \(\tan^{-1}7\) north of west.
