Statics and Dynamics Quiz: Cable Systems

What this quiz covers

This quiz focuses on Cable 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. $T_1 = 10\sqrt{3} \approx 17.3 \text{ kN}$ and $T_2 = 10 \text{ kN}$, found by resolving horizontal and vertical equilibrium with the sine rule applied to the triangle of forces.
2. $T_1 = 10 \text{ kN}$ and $T_2 = 10\sqrt{3} \approx 17.3 \text{ kN}$, found by resolving horizontal and vertical equilibrium with the sine rule applied to the triangle of forces. (correct answer)
3. $T_1 = 20/\sin 30° = 40 \text{ kN}$ and $T_2 = 20/\sin 60° \approx 23.1 \text{ kN}$, found by projecting each cable tension directly onto the vertical axis and setting each equal to $W$.
4. $T_1 = 20\cos 60° = 10 \text{ kN}$ and $T_2 = 20\cos 30° \approx 17.3 \text{ kN}$, found by decomposing $W$ into components along each cable direction using cosine projections.

Question 2

1. The system is statically indeterminate to the second degree because the three cables introduce three unknown tension forces and rigid-body equilibrium in 2D provides only two scalar equations, leaving two degrees of indeterminacy regardless of symmetry.
2. The system is statically indeterminate to the first degree. After applying symmetry ($T_2 = T_3$), two unknowns remain but only one independent equilibrium equation exists, because the moment equation is automatically satisfied for any cable tension values when the geometry is symmetric.
3. The system is statically determinate. Symmetry forces $T_2 = T_3$, reducing the problem to two unknowns. The vertical force equation ΣFy = 0 and the moment equation ΣM = 0 about the left edge then provide two independent equations sufficient to uniquely solve for $T_1$ and $T_2$. (correct answer)
4. The system is statically determinate because the horizontal equilibrium equation ΣFx = 0 provides an additional independent equation beyond ΣFy = 0 and ΣM = 0, yielding three equations for three unknowns and a unique solution.

Question 3

1. The reasoning is correct only when the external force $\mathbf{F}$ has no component along the cable direction; when $\mathbf{F}$ has a component along the cable axis, the cable force cannot be determined by inspecting the residual from the bars alone.
2. The reasoning is always correct for any joint configuration, because a cable can only pull along its axis; if the remaining members produce a residual force in the cable's compression direction, the cable is necessarily slack regardless of how many bars are present.
3. The reasoning is never correct at a joint with four or more members, because such joints are always statically indeterminate and the cable force cannot be isolated without invoking compatibility equations involving member stiffnesses.
4. The reasoning is correct only if the three rigid bars are statically determinate on their own — that is, the three bar forces can be uniquely found from the two equilibrium equations at P without the cable — and the resulting net residual force at P points in the direction the cable would need to push (compress) rather than pull. (correct answer)

Question 4

1. The student ignored the pin reaction at A. Taking moments about A: $T\sin35° \times 4 = 24 \times 6$, giving $T = 62.7 \text{ kN}$. The pin at A carries the remaining vertical and horizontal forces. (correct answer)
2. The student forgot to include the horizontal component of $T$ in the vertical equilibrium equation. The correct equation is $T\sin35° + T\cos35° = 24$, giving $T = 14.0 \text{ kN}$, with the horizontal component adding to the vertical equilibrium.
3. The student applied force equilibrium instead of moment equilibrium, and also used the wrong angle. The perpendicular distance from A to the cable line of action must be computed using $\cos35°$, giving $T = 24 \times 6/(4\cos35°) = 44.0 \text{ kN}$.
4. The student correctly identified the vertical equilibrium equation but applied it to the wrong free body. The full vertical equilibrium of the beam is $A_y + T\sin35° = 24$; since $A_y = 0$ for a pin support on a horizontal beam with no vertical load at A, the student's answer of 41.8 kN is actually correct.

Question 5

1. $W = H \times (\tan\theta_L + \tan\theta_R)$ where $\theta_L$ and $\theta_R$ are the slopes of the left and right cable segments at the low point; numerically $W = 50(4/10 + 4/20) = 50(0.4 + 0.2) = 30 \text{ kN}$. (correct answer)
2. $W = H \times 2 \times \tan\theta$, where $\theta$ is computed using the midspan geometry; numerically $W = 50 \times 2 \times (4/15) \approx 26.7 \text{ kN}$, since the lowest point must be at midspan for a single concentrated load.
3. $W = H \times \tan\theta_L$ alone, since at the lowest point the right cable segment is horizontal and contributes no vertical force; numerically $W = 50 \times (4/10) = 20 \text{ kN}$.
4. $W = 2H \times \sin\theta$, where $\theta$ is the angle of either cable segment; numerically $W = 2 \times 50 \times \sin(\arctan(4/15)) \approx 25.9 \text{ kN}$, since by symmetry both cable segments contribute equally to vertical equilibrium.