1. Which statement best describes force in elementary mechanics?
ⓐ. Force is stored in a moving body and is gradually used up as the body moves.
ⓑ. Force is a vector interaction that can change the state of motion of a body.
ⓒ. Force is an interaction that can occur only when two bodies are in direct contact.
ⓓ. Force is a vector that must always point in the same direction as the body’s velocity.
Correct Answer: Force is a vector interaction that can change the state of motion of a body.
Explanation: A force is associated with an interaction between a body and some external agency or another body. It is not a substance or property stored inside a moving object. A force has both magnitude and direction, so forces must be treated vectorially. An interaction may be a contact interaction, such as a push or tension, or it may act from a distance, as gravity does. A force can change the magnitude of velocity, its direction, or both. The direction of a force need not coincide with the direction in which the body is moving. A body may even continue moving when the force that initially changed its motion is no longer present. Thus the essential description is that force represents an interaction capable of affecting motion and is a vector quantity.
2. A stone is falling freely near Earth, a hand is pushing a box, and a taut string is pulling a suspended mass. Which classification of the corresponding interactions is correct?
ⓐ. All three forces are contact forces.
ⓑ. Gravitational force is non-contact, whereas the hand force and string tension are contact forces.
ⓒ. Gravitational force and tension are non-contact, whereas the hand force is a contact force.
ⓓ. Only the hand force is an interaction; gravity and tension are properties of the objects.
Correct Answer: Gravitational force is non-contact, whereas the hand force and string tension are contact forces.
Explanation: The falling stone interacts gravitationally with Earth even though there is no physical contact between them. Gravity is therefore a non-contact force in elementary mechanics. The hand can push the box only through direct contact with it, so the force exerted by the hand is a contact force. Tension in a string is transmitted through the material string and acts on the attached body at the point of contact. Hence tension is also treated as a contact force. Whether the body is moving or at rest does not determine whether the interaction is contact or non-contact. The classification depends on the physical interaction producing the force. It is also incorrect to regard tension or gravity as properties stored in the affected body. Therefore the gravitational interaction is non-contact, while the other two stated interactions are contact interactions.
3. Three horizontal forces act simultaneously on a small body: \(6\,\text{N}\) east, \(8\,\text{N}\) north and \(6\,\text{N}\) west. What is the net force on the body?
ⓐ. \(20\,\text{N}\) north-east
ⓑ. \(10\,\text{N}\) north-east
ⓒ. \(8\,\text{N}\) north
ⓓ. \(6\,\text{N}\) east
Correct Answer: \(8\,\text{N}\) north
Explanation: \( \textbf{Given:} \) The forces are \(6\,\text{N}\) east, \(8\,\text{N}\) north and \(6\,\text{N}\) west.
\( \textbf{Required:} \) Find their vector resultant.
\( \textbf{Choose components:} \) Take east as \(+x\) and north as \(+y\).
The horizontal resultant is
\[
F_x=6-6=0\,\text{N}.
\]
The vertical resultant is
\[
F_y=8\,\text{N}.
\]
Therefore,
\[
\vec F_{\text{net}}=8\,\hat j\,\text{N}.
\]
Its magnitude is \(8\,\text{N}\), and its direction is north. The two \(6\,\text{N}\) forces cancel because they are equal and opposite forces acting on the same body. Adding the three magnitudes arithmetically would ignore their directions and would therefore be incorrect. The net force is \(8\,\text{N}\) north.
4. Use the arrangement described below.
Two blocks \(P\) and \(Q\) are in contact on a smooth horizontal floor. A horizontal external force \(F\) is applied to block \(P\) towards the right. The selected system consists of both blocks \(P+Q\).
Which force must be treated as internal and therefore excluded from the net external force on the selected \(P+Q\) system?
ⓐ. The gravitational force exerted by Earth on block \(P\)
ⓑ. The normal force exerted by the floor on block \(Q\)
ⓒ. The applied force \(F\) on block \(P\)
ⓓ. The contact force between blocks \(P\) and \(Q\)
Correct Answer: The contact force between blocks \(P\) and \(Q\)
Explanation: The classification of a force as internal or external depends on the system chosen. Here the selected system contains both \(P\) and \(Q\). The force exerted by \(P\) on \(Q\) is therefore an interaction between two parts of the same system. Likewise, the force exerted by \(Q\) on \(P\) is also internal to the combined system. These mutual contact forces are not included when the net external force on \(P+Q\) is formed. The applied force \(F\) is external because the agent producing it lies outside the chosen two-block system. The gravitational forces are external because they are exerted by Earth, which is not part of the selected system. The floor normals are also external interactions with the floor. If block \(P\) alone were selected instead, the contact force exerted by \(Q\) on \(P\) would become external to that smaller system. Thus system choice, rather than the name of the force, determines whether an interaction is internal or external.
5. Use the arrangement described below.
A lamp is attached to a string fixed to the ceiling. A horizontal force exerted by a hand holds the lamp displaced to the right so that it remains at rest. The string therefore slopes upward and to the left from the lamp towards the ceiling support.
Which list correctly contains the forces acting on the lamp?
ⓐ. Weight downward, tension towards the support, the hand force to the right, and the reaction force exerted by the lamp on the string
ⓑ. Weight downward, tension along the string away from the ceiling support, and the hand force horizontally to the right
ⓒ. Weight downward, tension along the string towards the ceiling support, and the hand force horizontally to the right
ⓓ. Weight downward and the hand force to the right only, because a stationary lamp has no tension
Correct Answer: Weight downward, tension along the string towards the ceiling support, and the hand force horizontally to the right
Explanation: A free-body description must contain only forces acting on the selected body, which here is the lamp. Earth exerts the lamp's weight vertically downward. A taut string can pull but cannot push the attached body, so its tension on the lamp acts along the string towards the ceiling support. The hand exerts the stated horizontal force towards the right. These three interactions constitute the force set on the lamp. The force exerted by the lamp on the string acts on the string, not on the lamp, so it does not belong in the lamp's free-body representation. The fact that the lamp is at rest does not mean that no forces act. Instead, the three forces have a zero vector resultant. Reversing the direction of tension would incorrectly make the string push the lamp away from the support. Correct force identification must therefore precede any equation involving the lamp.
6. A block of mass \(m\) remains at rest on a horizontal table. In Case I, an additional vertical force \(P\) pushes the block downward. In Case II, the same magnitude \(P\), with \(P\lt mg\), pulls the block vertically upward without lifting it from the table. If the normal reactions are \(N_1\) and \(N_2\), respectively, what is \(N_1-N_2\)?
ⓐ. \(0\)
ⓑ. \(P\)
ⓒ. \(2P\)
ⓓ. \(2mg\)
Correct Answer: \(2P\)
Explanation: \( \textbf{Given:} \) The block remains at rest in both cases, so its vertical acceleration is zero.
\( \textbf{Case I:} \) The downward forces are \(mg\) and \(P\), while the normal reaction \(N_1\) acts upward.
Hence vertical force balance gives
\[
N_1-mg-P=0,
\]
so
\[
N_1=mg+P.
\]
\( \textbf{Case II:} \) The normal reaction \(N_2\) and applied force \(P\) act upward, while \(mg\) acts downward.
Therefore,
\[
N_2+P-mg=0,
\]
which gives
\[
N_2=mg-P.
\]
Thus,
\[
N_1-N_2=(mg+P)-(mg-P)=2P.
\]
The condition \(P\lt mg\) ensures that contact with the table can be maintained in Case II. This also demonstrates why the normal reaction is not automatically equal to \(mg\); it adjusts according to the other forces normal to the surface. Hence \(N_1-N_2=2P\).
7. The following bodies are observed at one instant.
| Body | Mass | Speed |
|---|
| \(X\) | \(2\,\text{kg}\) | \(10\,\text{m s}^{-1}\) |
| \(Y\) | \(5\,\text{kg}\) | \(0\) |
| \(Z\) | \(3\,\text{kg}\) | \(20\,\text{m s}^{-1}\) |
What is the correct decreasing order of their inertia?
ⓐ. \(Z\gt Y\gt X\)
ⓑ. \(Y\gt Z\gt X\)
ⓒ. \(Y\gt X\gt Z\)
ⓓ. \(Z\gt X\gt Y\)
Correct Answer: \(Y\gt Z\gt X\)
Explanation: In translational mechanics, mass is the measure of inertia. Inertia describes resistance to a change in the state of motion; it is not determined by how fast a body happens to be moving. Body \(Y\) has the greatest mass, \(5\,\text{kg}\), so it has the greatest inertia even though it is presently at rest. Body \(Z\), with mass \(3\,\text{kg}\), has the next greatest inertia. Body \(X\), with mass \(2\,\text{kg}\), has the least inertia of the three. The large speed of \(Z\) does not make its inertia exceed that of the more massive body \(Y\). Speed would matter when comparing quantities such as momentum, but inertia and momentum are not interchangeable ideas. A stationary body possesses inertia just as a moving body does. Therefore the decreasing order of inertia is \(Y\gt Z\gt X\).
8. Assertion: When a stationary bus suddenly starts moving forward, a standing passenger tends to fall backward relative to the bus.
Reason: The passenger's feet are accelerated forward by friction from the floor, while the upper part of the body initially tends to retain its state of rest because of inertia.
ⓐ. 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 of Assertion.
ⓒ. 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: The Assertion describes the passenger's motion relative to the accelerating bus. Before the bus starts, both the passenger and the bus are at rest relative to the ground. When the floor begins moving forward, friction at the passenger's feet provides the horizontal interaction that starts accelerating the feet forward. The rest of the body cannot acquire the same forward velocity instantaneously. Because of inertia, the upper body tends initially to preserve its previous state of rest. Consequently, relative to the forward-moving bus, the upper body appears to move backward. The passenger is not acted on by a mysterious backward force merely because the bus starts. The apparent backward tendency arises from the different immediate responses of the feet and upper body. Thus the Reason is true and directly explains the Assertion.
9. Consider the four situations below.
| Case | Information |
|---|
| P | A body is at rest while \(5\,\text{N}\) east and \(5\,\text{N}\) west act on it. |
| Q | A body moves east with constant velocity while \(3\,\text{N}\) east and \(3\,\text{N}\) west act on it. |
| R | A body is momentarily at rest while a single \(4\,\text{N}\) force acts northward. |
| S | A body moves with constant speed along a circular path. |
Which cases are consistent with zero acceleration at the stated instant or interval?
ⓐ. P, R and S only
ⓑ. P, Q and R only
ⓒ. Q and S only
ⓓ. P and Q only
Correct Answer: P and Q only
Explanation: In Case P, the two forces are equal and opposite on the same body, so their vector sum is zero. The stated rest condition is therefore consistent with zero acceleration. In Case Q, the forces again cancel, and constant velocity already means that both the magnitude and direction of velocity remain unchanged. Hence the acceleration is zero in Q as well. In Case R, being momentarily at rest tells us only that the instantaneous velocity is zero. The unbalanced northward force means the motion is being changed, so zero velocity at that instant must not be confused with zero acceleration. In Case S, the speed is constant but the direction of velocity continuously changes. A change in the velocity vector means non-zero acceleration, so a zero resultant is not possible for the circular motion described. The decisive distinction is therefore between zero velocity, constant velocity and constant speed with changing direction. Only P and Q are consistent with zero acceleration.
10. A puck is initially at rest on a frictionless horizontal surface. A horizontal force acts eastward for some time and is then removed. The puck subsequently moves with no horizontal force for an interval. Later, a westward force is applied until the puck's velocity becomes exactly zero, and this force is removed at that instant. What happens immediately afterward, assuming no other horizontal force acts?
ⓐ. It begins moving west because the last applied force was westward.
ⓑ. It begins moving east because it retains the effect of the first force.
ⓒ. It remains at rest until another horizontal force acts.
ⓓ. It alternately moves east and west because the earlier forces remain stored in it.
Correct Answer: It remains at rest until another horizontal force acts.
Explanation: During the first interval, the eastward force changes the puck's velocity from zero to an eastward value. Once that force is removed, it does not remain stored in the puck. With no horizontal resultant, the acquired velocity persists unchanged during the force-free interval. The later westward force changes that eastward velocity and is maintained until the instantaneous velocity becomes exactly zero. At that instant the westward force is also removed. The puck now has zero velocity and zero horizontal net force. Newton's first law then implies zero acceleration, so its zero velocity remains unchanged. The direction of a force that acted earlier does not determine the acceleration after that force has ceased. Similarly, the earlier eastward force cannot continue driving the puck after the interaction ends. Hence the puck remains at rest until a new resultant horizontal force acts.
11. Two blocks \(A\) and \(B\) remain in contact while moving to the right with constant velocity on a horizontal floor. A horizontal force \(P\) pushes block \(A\) to the right. The floor exerts frictional forces of magnitudes \(f_A\) and \(f_B\) to the left on the two blocks. If \(A+B\) is chosen as one system, which horizontal relation is correct?
ⓐ. \(P-f_A-f_B-F_{AB}=0\), where \(F_{AB}\) is the contact force between the blocks
ⓑ. \(P-f_A-f_B=0\)
ⓒ. \(P+f_A+f_B=0\)
ⓓ. \(P-F_{AB}=0\)
Correct Answer: \(P-f_A-f_B=0\)
Explanation: The two blocks are moving with constant velocity, so the acceleration of the combined system is zero. Newton's first-law condition therefore requires the net external horizontal force on the selected \(A+B\) system to vanish. The applied push \(P\) acts externally towards the right. The floor exerts the two external frictional forces \(f_A\) and \(f_B\) towards the left. The contact forces between \(A\) and \(B\) form an internal interaction when both blocks are included in the chosen system. Therefore that contact interaction must not appear in the whole-system external-force equation. Taking right as positive gives
\[
P-f_A-f_B=0.
\]
The equation also shows that a non-zero applied force can coexist with constant velocity when another external force balances it. Constant velocity does not require the absence of every force; it requires zero resultant external force. Hence the correct whole-system relation is \(P-f_A-f_B=0\).
12. Consider the following statements.
Statement I: If the velocity of a body is zero at a particular instant, the net external force on it must also be zero at that instant.
Statement II: If the net external force on a body remains zero over a time interval, its velocity remains constant over that interval, and that constant velocity may be zero or non-zero.
ⓐ. Both Statement I and Statement II are correct.
ⓑ. Both Statement I and Statement II are incorrect.
ⓒ. Statement I is correct but Statement II is incorrect.
ⓓ. Statement I is incorrect but Statement II is correct.
Correct Answer: Statement I is incorrect but Statement II is correct.
Explanation: Statement I incorrectly identifies instantaneous zero velocity with zero acceleration. A body can be momentarily at rest while its velocity is changing, in which case a non-zero resultant force may act. For example, the decisive issue in the first law is whether the velocity vector is constant, not whether its value happens to be zero at one instant. Therefore Statement I is incorrect. Statement II expresses the zero-net-force consequence of Newton's first law. When the resultant external force remains zero, acceleration is zero. Zero acceleration means that the velocity vector does not change with time. That constant velocity can be \(0\), corresponding to continued rest, or any constant non-zero vector, corresponding to uniform straight-line motion. Rest and uniform linear motion are therefore equivalent zero-acceleration states for this purpose. Hence Statement II is correct while Statement I is not.
13. Read the situation below and answer the question.
A trolley is pulled along a rough horizontal track by a constant horizontal force and initially moves with constant velocity. Without changing the pulling force, it then enters an ideal frictionless section of track. After moving some distance on this frictionless section, the pulling force is completely switched off.
Which description of the trolley's motion is correct?
ⓐ. Constant velocity throughout all three stages because the trolley was already moving before reaching the frictionless section
ⓑ. Constant velocity on the rough section, constant velocity on the frictionless section while the pull acts, then gradual slowing after the pull is removed
ⓒ. Increasing speed on the rough section, constant velocity while the pull acts on the frictionless section, then rest after the pull is removed
ⓓ. Constant velocity on the rough section, increasing speed while the pull acts on the frictionless section, then constant velocity after the pull is removed
Correct Answer: Constant velocity on the rough section, increasing speed while the pull acts on the frictionless section, then constant velocity after the pull is removed
Explanation: On the initial rough section, the trolley is stated to move with constant velocity. Therefore its acceleration is zero and the horizontal resultant must be zero. The applied pull is consequently balanced by the opposing frictional force during that stage. When the trolley enters the frictionless section, the opposing frictional force disappears but the pulling force is deliberately kept unchanged. The previously balanced force is now an unbalanced forward external force, so the trolley's velocity must begin changing and its speed increases. The fact that it had constant velocity before entering the smooth region does not guarantee constant velocity under the new force condition. Later the pulling force is switched off while the trolley is already moving faster than before. With the surface frictionless and the pull absent, the horizontal resultant then becomes zero. Newton's first law requires the velocity acquired by that instant to persist unchanged. Thus the correct sequence is constant velocity, then increasing speed, then a new constant velocity.
14. A block remains at rest on a rough horizontal floor while a string pulls it with tension \(T\) at an angle \(\theta\) above the horizontal towards the right. The block does not lose contact with the floor. Which pair correctly gives the normal reaction \(N\) and the frictional force \(f\) on the block?
ⓐ. \(N=mg+T\sin\theta\), \(f=T\cos\theta\) towards the right
ⓑ. \(N=mg-T\sin\theta\), \(f=T\sin\theta\) towards the left
ⓒ. \(N=mg-T\sin\theta\), \(f=T\cos\theta\) towards the left
ⓓ. \(N=mg\), \(f=T\cos\theta\) towards the left
Correct Answer: \(N=mg-T\sin\theta\), \(f=T\cos\theta\) towards the left
Explanation: \( \textbf{Force identification:} \) The block has weight \(mg\) downward, normal reaction \(N\) upward, tension along the string, and friction along the floor.
Resolve the tension into components:
\[
T_x=T\cos\theta,\qquad T_y=T\sin\theta.
\]
The horizontal component \(T\cos\theta\) tends to move the block to the right. Because the block remains at rest, friction must oppose this impending relative motion and therefore acts to the left. Horizontal balance gives
\[
f=T\cos\theta.
\]
Vertically, the upward forces are \(N\) and \(T\sin\theta\), while the downward force is \(mg\). Zero vertical acceleration requires
\[
N+T\sin\theta-mg=0.
\]
Hence
\[
N=mg-T\sin\theta.
\]
The upward component of tension therefore reduces the normal reaction below \(mg\). This result also shows why assuming \(N=mg\) automatically in every horizontal-surface problem is incorrect. Thus \(N=mg-T\sin\theta\) and friction has magnitude \(T\cos\theta\) towards the left.
15. A small puck moves in a circle on a smooth horizontal table because a horizontal string attached to its centre continuously pulls it towards a fixed point. At one instant the string breaks. Neglect all other horizontal forces. Which motion follows immediately after the break?
ⓐ. The puck moves radially outward with increasing speed.
ⓑ. The puck moves along the tangent at the breaking point with constant velocity.
ⓒ. The puck continues along the same circle but with decreasing speed.
ⓓ. The puck moves towards the centre and then comes to rest.
Correct Answer: The puck moves along the tangent at the breaking point with constant velocity.
Explanation: Before the string breaks, the puck's speed may be constant, but its velocity is not constant because its direction continually changes. The string supplies the inward interaction responsible for that change in direction. The instantaneous velocity of the puck at every point of the circular path is tangential to the circle. When the string breaks, its inward force disappears immediately. Under the stated condition, no other horizontal force remains, so the horizontal net force becomes zero. The velocity already possessed at the instant of breaking does not disappear with the force. Newton's first law therefore makes that instantaneous tangential velocity persist as a constant vector. There is no outward force that suddenly appears when the string breaks. The common impression that the puck should fly radially outward confuses its departure from the circle with the actual direction of its velocity. Hence it travels in a straight tangent with constant velocity.
16. A puck moves eastward on a frictionless horizontal surface. A brief northward push acts on it and is then completely removed. At the instant the push ends, the puck has both an eastward and a northward component of velocity. Which statement correctly describes its subsequent motion?
ⓐ. It moves north because the last force acted northward.
ⓑ. It gradually turns back towards the east because its original motion is restored.
ⓒ. It follows a curved path because the earlier northward force continues to influence its direction.
ⓓ. It moves in a straight line with the constant velocity vector it has at the instant the push ends.
Correct Answer: It moves in a straight line with the constant velocity vector it has at the instant the push ends.
Explanation: The brief northward interaction changes the puck's velocity while that interaction is present. By the instant the push ends, the puck has the stated eastward and northward velocity components. Those components combine vectorially to form its instantaneous resultant velocity. Once the push has ended, the puck does not carry a residual northward force with it. The surface is frictionless and no other horizontal force is specified, so the net horizontal force after that instant is zero. Newton's first law then requires the complete velocity vector to remain constant. Therefore both its magnitude and direction remain unchanged after the force is removed. The puck consequently follows a straight line in the direction of that resultant velocity. Its direction of motion need not coincide with the direction of any force that acted earlier. This is a direct example of the distinction between a body's present velocity and its present force: velocity can persist after the interaction that changed it has ceased.
17. A particle is acted upon by four coplanar forces: \(6\,\text{N}\) east, \(8\,\text{N}\) north, \(10\,\text{N}\) directed south-west with components \(6\,\text{N}\) west and \(8\,\text{N}\) south, and \(4\,\text{N}\) east. The \(10\,\text{N}\) force is then removed, while the \(4\,\text{N}\) eastward force is simultaneously reversed. What is the new resultant force?
ⓐ. \(10\,\text{N}\) towards north-east
ⓑ. \(2\sqrt{17}\,\text{N}\) towards south-east
ⓒ. \(2\sqrt{17}\,\text{N}\) towards north-east
ⓓ. \(8\,\text{N}\) towards north
Correct Answer: \(2\sqrt{17}\,\text{N}\) towards north-east
Explanation: \( \textbf{Given after the change:} \) The \(10\,\text{N}\) south-west force is absent and the \(4\,\text{N}\) force now acts westward.
Take east as \(+x\) and north as \(+y\).
The remaining horizontal components are
\[
F_x=6-4=2\,\text{N}.
\]
The remaining vertical component is
\[
F_y=8\,\text{N}.
\]
Therefore the magnitude of the new resultant is
\[
F_{\text{net}}=\sqrt{F_x^2+F_y^2}
=\sqrt{2^2+8^2}
=\sqrt{68}
=2\sqrt{17}\,\text{N}.
\]
Both components are positive, so the resultant points towards north-east. Its direction also satisfies \(\tan\theta=8/2=4\), measured north of east. The important reasoning step is to rebuild the force set after both changes rather than merely modify the magnitude of the original resultant. Hence the new resultant is \(2\sqrt{17}\,\text{N}\) towards north-east.
18. A small iron block rests on a smooth horizontal table. A magnet placed to the right of the block attracts it horizontally towards the right. Immediately before the magnet is removed, and immediately after it is removed, which description of the force sets on the block is correct?
ⓐ. Before removal: weight, normal reaction and magnetic force; after removal: weight and normal reaction only
ⓑ. Before removal: weight and magnetic force only; after removal: weight only
ⓒ. Before removal: weight, normal reaction and a force of motion to the right; after removal: the force of motion remains
ⓓ. Before removal: weight, normal reaction and magnetic force; after removal: magnetic force is replaced by friction
Correct Answer: Before removal: weight, normal reaction and magnetic force; after removal: weight and normal reaction only
Explanation: The block interacts gravitationally with Earth, so its weight acts vertically downward in both situations. The table is in contact with the block, so it exerts an upward normal reaction. Because the table is explicitly smooth, no frictional force is introduced. Before removal, the magnet interacts with the block without physical contact and exerts a horizontal magnetic force towards the magnet. Once the magnet is removed, that magnetic interaction no longer exists, so the magnetic force disappears from the block's free-body description. Removing the magnet does not remove the block's weight or the table contact. Motion acquired earlier is not itself an additional force and should never be drawn as a "force of motion." Nor does friction appear merely because another interaction has vanished. Thus the correct force set changes from three forces to weight plus normal reaction only.
19. Three blocks \(A\), \(B\) and \(C\) lie in a line on a smooth horizontal floor. \(A\) and \(B\) are connected by string \(S_1\), while \(B\) and \(C\) are connected by string \(S_2\). An external horizontal pull \(F\) acts on \(C\) towards the right. Which statement correctly compares the force classification for the systems \(A+B\) and \(A+B+C\)?
ⓐ. For \(A+B\), \(S_1\) is external and \(S_2\) is internal; for \(A+B+C\), \(S_1\) is external and \(S_2\) is internal.
ⓑ. For \(A+B\), both string interactions are internal and \(F\) is external; for \(A+B+C\), \(S_1\) is internal and \(S_2\) is external.
ⓒ. For \(A+B\), \(S_1\) is internal and \(S_2\) is external; for \(A+B+C\), \(S_1\) is external and \(S_2\) is internal.
ⓓ. For \(A+B\), \(S_1\) is internal and \(S_2\) is external; for \(A+B+C\), both string interactions are internal and \(F\) is external.
Correct Answer: For \(A+B\), \(S_1\) is internal and \(S_2\) is external; for \(A+B+C\), both string interactions are internal and \(F\) is external.
Explanation: Whether a force is internal or external cannot be decided without first choosing the system. For the system \(A+B\), both ends of the \(S_1\) interaction involve bodies inside the selected system, so that interaction is internal. Block \(C\), however, lies outside this system. Therefore the pull exerted through \(S_2\) on block \(B\) crosses the boundary of the \(A+B\) system and is an external horizontal force on it. The applied force \(F\) acts directly on \(C\), which is not part of the \(A+B\) system, so \(F\) itself is not drawn directly on the free-body representation of \(A+B\). If all three blocks are instead chosen as one system, both connecting-string interactions occur entirely within the system. They are therefore internal interactions for \(A+B+C\). The externally applied pull \(F\) now acts directly on a member of the chosen system and is external to it. This illustrates why system selection must be made before writing a net-external-force equation.
20. Three forces acting on a particle initially give zero resultant. Two of them are \(15\,\text{N}\) east and \(8\,\text{N}\) north, while the third force has the required magnitude and direction to balance them. If only the northward force is changed from \(8\,\text{N}\) to \(16\,\text{N}\), with the other two forces unchanged, what is the new resultant?
ⓐ. \(15\,\text{N}\) east
ⓑ. \(8\,\text{N}\) north
ⓒ. \(17\,\text{N}\) north-east
ⓓ. \(8\,\text{N}\) south
Correct Answer: \(8\,\text{N}\) north
Explanation: \( \textbf{Initial condition:} \) The resultant of all three original forces is zero.
The resultant of the first two forces is
\[
\vec R=(15\,\hat i+8\,\hat j)\,\text{N}.
\]
Its magnitude is
\[
R=\sqrt{15^2+8^2}=17\,\text{N}.
\]
Therefore the third force must initially be
\[
\vec F_3=-(15\,\hat i+8\,\hat j)\,\text{N}.
\]
After the change, the northward force becomes \(16\,\hat j\,\text{N}\), while the eastward and balancing forces are unchanged. Hence
\[
\vec F_{\text{new}}
=15\hat i+16\hat j-15\hat i-8\hat j
=8\hat j\,\text{N}.
\]
The horizontal components still cancel completely. Only an additional \(8\,\text{N}\) northward component remains. Thus the new resultant is \(8\,\text{N}\) north.
