Statics and Dynamics Quiz: Concurrent Force Systems

What this quiz covers

This quiz focuses on Concurrent Force Systems, 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. $P : R = 1 : 2$, because $R$ acts closer to the negative x-axis than $Q$, so its x-component is larger in magnitude than P's, requiring R to be twice P to balance the combined x-projections of $Q$ and $P$ in the negative direction.
2. $P : R = \sqrt{3} : 1$, because the horizontal equilibrium equation couples P and R through their cosine projections in a way that the x-component of Q creates an asymmetry between P and R that scales as $\sqrt{3}$.
3. $P : R = 2 : 1$, because resolving equilibrium shows the x-component of Q partially cancels R's x-component, leaving P to carry twice the horizontal load that R provides, independent of Q's magnitude.
4. $P : R = 1 : 1$, because for any symmetric three-force concurrent system with one force at 120° and another at 240°, all three magnitudes must be equal to satisfy both horizontal and vertical equilibrium simultaneously. (correct answer)

Question 2

1. $\mathbf{F}_6 = (-1\hat{i} + 3\hat{j}) \text{ kN}$ with magnitude $\sqrt{10} \approx 3.16 \text{ kN}$, because restoring equilibrium after adding $\mathbf{F}_5$ requires only canceling $\mathbf{F}_5$ itself, since the original four forces were already balanced. (correct answer)
2. $\mathbf{F}_6 = (1\hat{i} + 3\hat{j}) \text{ kN}$ with magnitude $\sqrt{10} \approx 3.16 \text{ kN}$, because $\mathbf{F}_6$ must match the y-component of $\mathbf{F}_5$ while reversing the x-component to maintain the original horizontal balance.
3. $\mathbf{F}_6 = (-1\hat{i} - 3\hat{j}) \text{ kN}$ with magnitude $\sqrt{10} \approx 3.16 \text{ kN}$, because the new equilibrant must point in the same direction as $\mathbf{F}_5$ to create a couple that neutralizes the added force without changing the net moment at P.
4. $\mathbf{F}_6 = (2\hat{i} - 6\hat{j}) \text{ kN}$ with magnitude $\sqrt{40} \approx 6.32 \text{ kN}$, because $\mathbf{F}_6$ must counteract both $\mathbf{F}_5$ and re-balance $\mathbf{F}_4$, which was determined under the old system and must be recomputed as twice the equilibrant of $\mathbf{F}_5$.

Question 3

1. The claim is correct, because for small angles $\sin\alpha \approx \alpha$ and $\cos\alpha \approx 1$, so the equilibrium equations become linear in $\alpha$, making $T_1$ directly proportional to $\alpha$ and thus doubling when $\alpha$ doubles.
2. The claim is incorrect, because $T_1$ depends on $\sin\alpha$ through the horizontal equilibrium equation and $\sin\alpha$ is not proportional to $\alpha$ in the relevant range; the relationship is always nonlinear regardless of angle magnitude.
3. The claim is incorrect, because even for small $\alpha$, the equilibrium equations couple $T_1$ and $T_2$ through both $\alpha$ and $\beta$; doubling $\alpha$ changes both $T_1$ and $T_2$, and the net effect on $T_1$ alone is not a simple doubling due to the coupled system of equations. (correct answer)
4. The claim is correct, because the vertical equilibrium equation shows $T_1 \cos\alpha + T_2 \cos\beta = W$, and for small $\alpha$, $\cos\alpha \approx 1$, making $T_1 \approx W - T_2$, which is independent of $\alpha$ and thus shows $T_1$ does not change, contradicting the doubling claim while confirming the approximation is valid.

Question 4

1. $T_A = \frac{W \cos50°}{\sin70°}$ and $T_B = \frac{W \cos20°}{\sin70°}$, where $W = 1962 \text{ N}$, because applying Lami's theorem to the three concurrent forces at the junction yields these expressions directly. (correct answer)
2. $T_A = \frac{W \sin50°}{\sin70°}$ and $T_B = \frac{W \sin20°}{\sin70°}$, where $W = 1962 \text{ N}$, because the equilibrium equations resolved along the cable directions use sine projections of the weight, not cosine projections.
3. $T_A = \frac{W}{\sin20° + \sin50°}$ and $T_B = \frac{W}{\sin20° + \sin50°}$, where $W = 1962 \text{ N}$, because the equal horizontal projections of the two tensions must sum to zero, making the tensions equal and each supporting half the load divided by the average sine.
4. $T_A = W\sin50°$ and $T_B = W\sin20°$, where $W = 1962 \text{ N}$, because the vertical components of each cable tension individually equal the weight projected onto that cable's direction, without needing simultaneous equations.

Question 5

1. $500\cos35° + F_{AB}\cos70° - F_{AC}\cos20° + N\cos105° = 0$ and $500\sin35° + F_{AB}\sin70° - F_{AC}\sin20° + N\sin105° = 0$, because the compression in AC means its contribution is subtracted using the original member angle of $-20°$ rather than the actual pin-force direction of $160°$.
2. $500\cos35° + F_{AB}\cos70° + F_{AC}\cos160° + N\cos105° = 0$ and $500\sin35° + F_{AB}\sin70° + F_{AC}\sin160° + N\sin105° = 0$, treating all forces with their actual direction angles from the positive x-axis so that all sign information is carried by the trigonometric functions. (correct answer)
3. $500\cos35° + F_{AB}\cos70° + F_{AC}\cos160° + N\cos105° = 0$ and $500\sin35° + F_{AB}\sin70° - F_{AC}\sin160° + N\sin105° = 0$, because compression in AC reverses the sign of its y-component only, since the vertical direction is where compression effects are felt at the pin.
4. $500\cos35° + F_{AB}\cos70° + F_{AC}\cos160° + N\cos90° = 0$ and $500\sin35° + F_{AB}\sin70° + F_{AC}\sin160° + N\sin90° = 0$, because the roller reaction is always perpendicular to the ground, and since perpendicular to horizontal means vertical, $N$ acts straight upward at $90°$ regardless of surface inclination.