Statics and Dynamics Quiz: Rigid Body Force Equation

What this quiz covers

This quiz focuses on Rigid Body Force Equation, 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. Required friction $f = mg\sin\theta/2 = 24.5 \text{ N}$; maximum static friction $= \mu_s mg\cos\theta = 33.97 \text{ N}$. Since the required friction is less than the maximum, the cylinder rolls without slipping and the correct acceleration is $a_G = \tfrac{1}{2}g\sin\theta = 2.45 \text{ m/s}^2$.
2. Required friction $f = mg\sin\theta/3 = 16.35 \text{ N}$; maximum static friction $= \mu_s mg\cos\theta = 33.97 \text{ N}$. Since the required friction is less than the maximum, the cylinder rolls without slipping and the correct acceleration is $a_G = \tfrac{2}{3}g\sin\theta = 3.27 \text{ m/s}^2$. (correct answer)
3. Required friction $f = mg\sin\theta/3 = 16.35 \text{ N}$; maximum static friction $= \mu_s mg = 39.24 \text{ N}$. Since the required friction is less than the maximum, the cylinder rolls without slipping and the correct acceleration is $a_G = \tfrac{2}{3}g\sin\theta = 3.27 \text{ m/s}^2$.
4. Required friction $f = mg\sin\theta/3 = 16.35 \text{ N}$; maximum static friction $= \mu_s mg\cos\theta = 33.97 \text{ N}$. Since the required friction is less than the maximum, the cylinder rolls without slipping and the correct acceleration is $a_G = \tfrac{1}{3}g\sin\theta = 1.63 \text{ m/s}^2$.

Question 2

1. $a_G = \tfrac{3}{4}g$, because the moment equation about the remaining support gives a constraint that, combined with $\sum F_y = M a_G$, yields the remaining cable tension as $Mg/4$, so the net downward force is $\tfrac{3}{4}Mg$. (correct answer)
2. $a_G = g$, because once a cable is cut the plate is in free fall and the remaining cable instantaneously loses tension, making the net force equal to the full weight $Mg$.
3. $a_G = \tfrac{1}{2}g$, because by symmetry the remaining cable still carries half the weight, leaving a net downward force of $Mg/2$ and thus $a_G = g/2$.
4. $a_G = \tfrac{1}{4}g$, because the remaining cable tension drops to $\tfrac{3}{4}Mg$ immediately after the cut, leaving a net downward force of only $Mg/4$ on the plate.

Question 3

1. $A_x = m(a_0 + \alpha L/2)$, because the weight $mg$ acts downward and produces a restoring torque, so the angular acceleration always assists the forward motion of $G$, making the tangential contribution add to $a_0$ regardless of the direction of $\alpha$.
2. $A_x = m a_0$, because the center of mass must have the same horizontal acceleration as the pin ($a_0$) since the bar is rigid and the pin constrains horizontal motion of end $A$, leaving no additional contribution from rotation.
3. $A_x = m(a_0 - \alpha L/2)$, because the center of mass acceleration is $a_{Gx} = a_0 + (\alpha_{rel})(L/2)$, where for a bar hanging vertically a counterclockwise $\alpha$ swings $G$ to the left relative to $A$, so the tangential contribution is $-\alpha L/2$ in the rightward-positive convention, giving $A_x = m(a_0 - \alpha L/2)$. (correct answer)
4. $A_x - mg = m a_{Gx}$, giving $A_x = m(g + a_0)$, because the horizontal equation must include the weight component and the inertial resistance to the imposed acceleration $a_0$ of the pin.

Question 4

1. The net force is zero and therefore the body must be either stationary or moving with constant velocity in a straight line, while angular velocity must also be constant because $\alpha$ is coupled to $a_G$ through the rigid-body kinematic equations.
2. Both the net force and the net moment about every point must be zero, because if $\mathbf{a}_G = 0$ the body is in static equilibrium and no angular acceleration is possible without a translational acceleration of $G$.
3. The net force on the body is zero ($\sum \mathbf{F} = 0$), but the net moment about $G$ may be nonzero, allowing the body to have a nonzero angular acceleration $\alpha$ about its center of mass at that instant. (correct answer)
4. The net force on the body is zero ($\sum \mathbf{F} = 0$), and consequently the net moment about any point other than $G$ must also be zero, because moment equilibrium at any off-center point would otherwise imply a nonzero $\sum \mathbf{F}$ by the moment-force relationship.

Question 5

1. The crate tips before sliding. The friction required to prevent sliding is $f = ma = 250 \text{ N}$, which is less than $\mu_s N = 294 \text{ N}$, so sliding does not occur. However, the tipping check (moment about the front bottom edge) shows the normal force would need to act at $x = h/(2) \cdot (a/g) - w/2 = 0.306 - 0.4 = -0.094 \text{ m}$ from center, which is within the base, so the crate does not tip either — it remains stationary.
2. The crate slides without tipping. The friction required is $f = ma = 250 \text{ N}$, which exceeds $\mu_s N = \mu_s mg = 0.6(50)(9.81) = 294 \text{ N}$, so the friction demand exceeds the maximum and sliding occurs before any tipping can develop.
3. The crate tips before sliding. The friction required is $f = ma = 250 \text{ N} < \mu_s mg = 294 \text{ N}$, so sliding is prevented. For the tipping check, the moment equation about the forward bottom edge gives the normal force location as $x = w/2 - (a/g)(h/2) = 0.4 - 0.306 = 0.094 \text{ m}$ from the forward edge, which is within the base (positive), confirming the crate does not tip and remains stationary.
4. The crate tips before sliding. Since $f = ma = 250 \text{ N} < \mu_s mg = 294 \text{ N}$, sliding is prevented. The tipping condition requires the normal force to reach the forward bottom edge; the critical deceleration for tipping is $a_{tip} = g(w/h) = 9.81(0.8/1.2) = 6.54 \text{ m/s}^2$. Since $a = 5 \text{ m/s}^2 < 6.54 \text{ m/s}^2$, tipping does not occur and the crate remains stationary. (correct answer)

Question 6

1. The error is treating the pulley as a particle; the correct approach replaces $M a_G$ with the effective mass $M_{eff} = M/2$ in a modified translational equation $T_B - T_A = M_{eff} \cdot a$, which is algebraically identical to the moment equation and yields $a = mg/(3m + M/2)$.
2. The error is using $M a_G$ instead of $M a_G/2$ for a rotating cylinder; the corrected translational equation for the pulley is $T_B - T_A = M a_G/2$, and since the pulley's center moves with acceleration $a$, the system acceleration is $a = mg/(2m + m + M/2)$.
3. The error is neglecting the bearing reaction at the pulley axle; once included, the net force on the pulley is zero regardless of $T_A$ and $T_B$, so $T_A = T_B$ and the system acceleration simplifies to $a = mg/(3m)= g/3$, independent of $M$.
4. The pulley's center does not translate, so $a_G = 0$ for the pulley and $\sum F = M a_G = 0$ is satisfied by bearing forces, not by $T_B - T_A$. The pulley requires a moment equation $\sum M_G = I_G \alpha$, giving $(T_B - T_A)R = (MR^2/2)\alpha = (MR^2/2)(a/R)$, so $T_B - T_A = Ma/2$. The correct acceleration is $a = mg/(2m + m + M/2)$. (correct answer)

Question 7

1. $\sum M_A = I_A \alpha$ is valid only when $A$ is the center of mass or a fixed pivot. For a general body-fixed point, the correct form is $\sum M_A = I_G \alpha + \mathbf{r}_{G/A} \times m\mathbf{a}_G$, which equals $I_A \alpha$ only in the special case that $\mathbf{a}_G$ is directed along $\mathbf{r}_{G/A}$ (i.e., the two vectors are parallel, so their cross product vanishes).
2. $\sum M_A = I_A \alpha$ is valid only when $A$ is the center of mass or a fixed pivot ($\mathbf{a}_A = 0$). The general correct form for any body-fixed point is $\sum M_A = I_A \alpha - m\,\mathbf{r}_{G/A} \times \mathbf{a}_A$, where the correction term accounts for the acceleration of point $A$; this reduces to $I_A \alpha$ precisely when $\mathbf{a}_A = 0$. (correct answer)
3. $\sum M_A = I_A \alpha$ is always valid for any body-fixed point $A$, because the parallel-axis theorem guarantees that $I_A$ already accounts for the geometric offset from $G$, making additional correction terms unnecessary regardless of how point $A$ is accelerating.
4. $\sum M_A = I_A \alpha$ is valid whenever $A$ is the instantaneous center of zero velocity, because at that instant $A$ momentarily acts as a fixed pivot and the standard fixed-point equation applies exactly, with no inertia correction required.

Question 8

1. $mg - A_y = m a_{Gy}$, with $a_{Gy} = \tfrac{3g}{4}$ (downward) from the rotational kinematics, giving $A_y = mg/4$ upward, meaning the pin supports only one-quarter of the rod's weight at the instant of release. (correct answer)
2. $mg - A_y = m a_{Gy}$, with $a_{Gy} = \tfrac{g}{2}$ (downward) because the center moves in a circular arc and the centripetal acceleration is zero at the instant of release while the tangential component equals $\alpha L/2 = g/2$, giving $A_y = mg/2$ upward.
3. $A_y - mg = m a_{Gy}$, with $a_{Gy} = \tfrac{3g}{4}$ (upward) because the pin must accelerate the rod upward to keep the pin end fixed, giving $A_y = mg + m(3g/4) = \tfrac{7mg}{4}$ upward.
4. $mg - A_y = m a_{Gy}$, with $a_{Gy} = 0$ because the rod starts from rest and has no velocity, so the center of mass has zero acceleration at the instant of release, giving $A_y = mg$ upward.

Question 9

1. $P - f = m a_G$, giving $a_G = 3.33 \text{ m/s}^2$, where $f$ is the friction force found from the rolling constraint and the moment equation to be $P/3 = 13.33 \text{ N}$. (correct answer)
2. $P = m a_G$, giving $a_G = 5.0 \text{ m/s}^2$, because friction acts perpendicular to motion for rolling contact and therefore does not appear in the horizontal force equation.
3. $P - f = m a_G$, giving $a_G = 1.67 \text{ m/s}^2$, where friction is taken as the maximum static value $f = \mu_s m g = 19.6 \text{ N}$ regardless of the rolling constraint.
4. $P + f = m a_G$, giving $a_G = 6.67 \text{ m/s}^2$, because the friction force at the contact point acts in the same direction as $P$ to oppose relative slip and thereby assists the forward motion of the center.