Statics and Dynamics Quiz: Dynamic Free Body Diagrams

What this quiz covers

This quiz focuses on Dynamic Free Body Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The cylinder's kinetic couple $I_G\alpha = mr^2\alpha$ is larger than the disk's $I_G\alpha = \frac{1}{2}mr^2\alpha$ for the same $\alpha$, but since the actual friction force is set by the surface condition $f = \mu_s mg\cos\theta$, the friction forces on both FBDs are equal regardless of the body's rotational inertia.
2. Both bodies have the same $ma$ term and the same $I_G\alpha$ couple on their kinetic diagrams because they share identical mass and radius; the differing accelerations arise only from the incline angle, which affects both bodies equally.
3. The cylinder has a larger $I_G\alpha$ couple than the disk for equal $\alpha$; applying the equations of motion with the rolling constraint shows that a larger rotational couple requires a larger friction force on the FBD, consistent with the cylinder having a lower linear acceleration down the incline. (correct answer)
4. The cylinder's kinetic couple is larger; solving $\sum F = ma$ and $\sum M_G = I_G\alpha$ with the rolling constraint yields $f = \frac{1}{3}mg\sin\theta$ for both bodies, so the friction magnitudes are equal even though the accelerations differ.

Question 2

1. Normal inertial term: $\frac{mv^2}{\rho}$ toward the center of curvature. Tangential inertial term: $m|\dot{v}|$ directed opposite to the direction of motion (i.e., in the $-t$ direction), because deceleration means $\dot{v} < 0$ and the inertial term must explicitly reflect the backward physical direction.
2. Normal inertial term: $\frac{mv^2}{\rho}$ directed away from the center of curvature (outward centrifugal direction). Tangential inertial term: $m|\dot{v}|$ directed opposite to motion. Both terms are treated as d'Alembert inertia forces opposing the acceleration.
3. Normal inertial term: $\frac{mv^2}{\rho}$ toward the center of curvature. Tangential inertial term: $m\dot{v}$ drawn in the $+t$ direction (the direction of motion), where $\dot{v}$ is negative for deceleration; the algebra then yields a negative value, indicating the term physically points backward — consistent with decelerating motion. (correct answer)
4. Normal inertial term: $\frac{mv^2}{\rho}$ toward the center of curvature. Tangential inertial term: zero, because the net tangential deceleration is produced entirely by the external forces shown on the FBD, and placing $m\dot{v}$ on the kinetic diagram would count that effect twice.

Question 3

1. The second student is correct: $F = ma_G$ fully determines the motion. The couple $I_G\alpha$ is only needed when writing the rotational scalar equation separately, and placing it on the kinetic diagram is redundant for bodies where a single force drives the motion.
2. The first student is correct: the kinetic diagram must always show both $ma_G$ at G and $I_G\alpha$ as a couple about G for any rigid body in general planar motion. The two terms serve separate equations ($\sum F = ma_G$ and $\sum M_G = I_G\alpha$); the second student confuses the kinetic diagram with a single scalar equation of motion. (correct answer)
3. The first student is correct in placing both terms, but the couple $I_G\alpha$ should be drawn about the contact point with the surface rather than about G, because the frictionless surface makes the contact point the natural moment center for the rotational equation.
4. The second student is correct: because the surface is frictionless, there is no torque about G from the surface, so the rotational inertia $I_G\alpha$ is zero and only the translational term $ma_G$ needs to appear on the kinetic diagram.

Question 4

1. Forces on FBD: weight $mg$ downward, normal force $N$ upward from cart, and friction $f$ to the right from cart. Kinetic diagram inertial term: $ma$ to the right at the block's center of mass, no couple because the block translates without rotation. (correct answer)
2. Forces on FBD: weight $mg$ downward, normal force $N$ upward from cart, friction $f$ to the right from cart, and a pseudo-force $ma$ to the left representing the inertia of the block. Kinetic diagram inertial term: zero, because the pseudo-force already accounts for dynamics.
3. Forces on FBD: weight $mg$ downward, normal force $N$ upward from cart, friction $f$ to the right from cart, and a normal force from the floor acting upward on the block. Kinetic diagram inertial term: $ma$ to the right at the block's center of mass.
4. Forces on FBD: weight $mg$ downward and normal force $N$ upward from the cart only, because friction is an internal force between the block-cart system. Kinetic diagram inertial term: $(m+M)a$ to the right representing the system's inertia.

Question 5

1. A single resultant force $m\bar{a}$ acting at the center of mass directed straight downward, plus a couple $I_G \alpha$ about the center of mass, where $\bar{a}$ has only a tangential component because $\omega = 0$. (correct answer)
2. A single resultant force $m\bar{a}$ acting at the center of mass directed straight downward, plus a couple $I_O \alpha$ about pin point O, because the parallel-axis theorem allows the rotational inertia to be expressed about any convenient point.
3. A force $m\bar{a}_t$ tangent to the path of the center of mass acting at the center of mass, with no normal component (since $\omega = 0$) and no couple term, because a slender rod's rotational inertia is negligible compared to its translational inertia.
4. A couple $I_G \alpha$ about the center of mass only, with no translational inertial force, because the pin at O fully constrains translation so the net translational acceleration of the center of mass is zero at the instant of release.

Question 6

1. The student's claim is correct: because wire B still holds end B fixed at this instant, the bar rotates about B, and $I_B\alpha$ is the legitimate single inertial term on the kinetic diagram. The error would only arise if the student tried to use $I_B\alpha$ after B begins to move.
2. The error is that wire B constrains only vertical translation at B, not horizontal translation, so B is not a truly fixed point. Because the bar does not rotate about a fixed axis, the kinetic diagram must show $m\bar{a}_G$ at G plus $I_G\alpha$ about G rather than any single term about B.
3. The error is a sign convention issue: $I_B\alpha$ is acceptable as the sole kinetic diagram term, but the student must use the parallel-axis theorem $I_B = I_G + m(L/2)^2$ and define $\alpha$ as positive clockwise; using counterclockwise positive produces an incorrect sign for the tension in wire B.
4. The error is that $I_B\alpha$ belongs in the moment equation, not on the kinetic diagram itself. The kinetic diagram must always show $m\bar{a}_G$ at G plus $I_G\alpha$ about G. Even for a true fixed pin at B, these two separate terms appear on the kinetic diagram; $I_B\alpha$ emerges only when the moment equation is summed about B algebraically. (correct answer)

Question 7

1. Rod BC has inertial terms $m\bar{a}_{G_{BC}}$ at its center of mass and couple $I_{G_{BC}}\alpha_{BC}$, where $\bar{a}_{G_{BC}}$ has only a downward tangential component (no normal component) because $\omega = 0$ at release, and $\alpha_{BC} \neq 0$ so the couple is nonzero.
2. Rod BC has inertial terms $m\bar{a}_{G_{BC}}$ at its center of mass and couple $I_{G_{BC}}\alpha_{BC}$, but since B itself is accelerating (not fixed), the acceleration of the center of mass of BC must account for both the motion of B and the rotation of BC relative to B, making $\bar{a}_{G_{BC}}$ different in magnitude and direction from that of a simple pendulum pinned at a fixed point. (correct answer)
3. Rod BC has only the couple $I_B\alpha_{BC}$ about pin B as its single inertial term, since B is the instantaneous center of rotation for BC at this instant, allowing the translational and rotational inertia to be combined into one moment-of-inertia term.
4. Rod BC has inertial terms $m\bar{a}_{G_{BC}}$ at its center of mass and couple $I_{G_{BC}}\alpha_{BC}$, and since B is a pin joint (not a fixed support), rod BC behaves identically to an independent rod pinned at a fixed point B, so $\bar{a}_{G_{BC}} = \frac{L}{2}\alpha_{BC}$ directed straight downward.

Question 8

1. Only the couple $I_G\alpha = 0$ appears on the right-hand side, because the crate translates without rotating ($\alpha = 0$), so the moment of the inertial terms about any point is zero and the right-hand side vanishes entirely.
2. The term $I_A \alpha$ appears on the right-hand side, where $I_A$ is the moment of inertia about corner A, because summing moments about A for a body that could potentially rotate about A requires using the moment of inertia about that point.
3. The terms $ma_G \cdot h$ and $ma_G \cdot w$ both appear on the right-hand side, representing the moments of the horizontal and vertical components of $m\bar{a}_G$ about corner A, because $m\bar{a}_G$ has both horizontal and vertical components due to the inclined applied force $P$.
4. The term $m a_G \cdot h$ appears on the right-hand side, representing the moment of the translational inertial force $ma_G$ (acting horizontally at G) about point A. The rotational term $I_G\alpha = 0$ vanishes, but the translational inertial force still has a nonzero moment arm about A equal to $h$. (correct answer)