Equilibrium of Rigid Bodies Questions and Answers

Answer: B

The tensions are T₁ = 0 N and T₂ = 100 N.

The bar is uniform, so its 100 N weight acts at its midpoint, and string S2 is attached exactly there. Take moments about the midpoint: T₂ has no moment arm, the weight has none either, and T₁ acts L/2 away. For rotational equilibrium T₁ × L/2 = 0, so T₁ = 0, and vertical balance then gives T₂ = 100 N.

Splitting the weight 75/25 or 25/75 would be right only if the strings were placed elsewhere, and T₁ = 100 N with T₂ = 0 would leave the weight with a moment about S1, so the bar would rotate.

Answer: A

The minimum impulse is 0.953 Ns.

The force pushes the block towards the pinned corner, so the block starts to rotate about the pin O and lifts off the roller. Two steps: the impulse sets the angular velocity, and that kinetic energy must lift the centre of mass over O.

Impulse to angular momentum about O (the pin reaction has no moment about O):

\[ \begin{aligned} I_{F}\,d &= J_{O}\,\omega \\ J_{O} &= \tfrac{m}{3}(h^{2} + b^{2}) \\ &= \tfrac{1}{3}(0.16 + 0.09) = 0.0833 \end{aligned} \]

Energy to topple. The centre of mass starts \(h/2 = 0.2\) m above O and must rise to directly above it, at the half-diagonal \(\sqrt{0.2^{2} + 0.15^{2}} = 0.25\) m:

\[ \begin{aligned} \tfrac{1}{2}J_{O}\omega^{2} &= mg(0.25 - 0.20) = 0.4905 \\ \omega^{2} &= 11.77, \quad \omega = 3.431\ \text{rad/s} \\ I_{F} &= \frac{0.0833 \times 3.431}{0.3} = 0.953\ \text{Ns} \end{aligned} \]

Using the centroidal moment of inertia \(\tfrac{m}{12}(h^{2} + b^{2})\) instead of the one about the pin, or taking the rise of the centre of mass as the full 0.25 m, gives values that do not match; the key is the parallel-axis value about O with a rise of only 0.05 m.

Answer: A

The vertical reaction at C is 5 kN.

The hinge transmits no moment, so treat the left member A-hinge first. It is 1 m long with the 10 kN load at its middle, so A and the hinge each carry 5 kN. The hinge therefore pushes 5 kN down on the right member.

The right member runs from the hinge to C (1.0 m), with roller B 0.2 m from the hinge and the 25 kN load 0.4 m from the hinge. Take moments about B:

\[ \begin{aligned} R_{C}(0.8) &= 25(0.2) - 5(0.2) = 4 \\ R_{C} &= 5\ \text{kN} \end{aligned} \]

Then \(R_{B} = 30 - 5 = 25\) kN. Answers of 10, 15 or 20 come from treating the beam as one continuous member, ignoring the hinge's zero-moment condition.

Answer: A

The natural frequency is 10 rad/s.

Treat the beam as a spring under the mass. For an overhanging beam with span \(L = 1\) m and overhang \(a = 2\) m, a tip load \(F\) deflects the tip by

\[ \begin{aligned} \delta &= \frac{Fa^{2}(L + a)}{3EI} = \frac{F(4)(3)}{3EI} = \frac{4F}{EI} \\ k &= \frac{EI}{4} = \frac{200\times 10^{9}\times 10^{-8}}{4} = 500\ \text{N/m} \\ \omega_{n} &= \sqrt{k/m} = \sqrt{500/5} = 10\ \text{rad/s} \end{aligned} \]

The first term of the deflection comes from the overhang bending as a cantilever and the rest from rotation over the roller; dropping either part gives a stiffer spring and a higher, wrong frequency. 5 and the fractional values do not follow from \(k = 500\) N/m.

Answer: A

The reaction is \(\frac{mg}{4}\) in the positive y-direction.

Just after the cut the rod is still at rest, so only the angular acceleration about O matters. Taking moments about the hinge:

\[ \begin{aligned} \frac{mL^{2}}{3}\alpha &= mg\frac{L}{2},\quad \alpha = \frac{3g}{2L} \\ a_{G} &= \alpha\frac{L}{2} = \frac{3g}{4}\ \text{(downward)} \end{aligned} \]

Newton's law for the vertical motion of the centre of mass gives \(mg - R = m\cdot 3g/4\), so \(R = mg/4\), upward. mg would be the static reaction with no thread, and the downward options have the wrong sense: the hinge must still support part of the weight.

Answer: D

The reactions are \(T_{x} = 0\), \(T_{y} = 500\) N and \(W_{y} = 1500\) N.

Treat the whole frame as one free body. W is a roller, so it gives only a vertical reaction; the only other horizontal force could come from the pin at T, and there is no horizontal load, so \(T_{x} = 0\).

Take moments about T. The 2000 N load at S acts 3.6 m to the right of the line PT, and W is 4.8 m from T:

\[ \begin{aligned} W_{y} \times 4.8 &= 2000 \times 3.6 \Rightarrow W_{y} = 1500\ \text{N} \\ T_{y} &= 2000 - 1500 = 500\ \text{N} \end{aligned} \]

The internal pins at P, Q and R do not affect the support reactions of the frame as a whole. Equal shares of 1000 N would need the load midway between T and W (2.4 m), and the 1500/500 split the other way round puts the load nearer T than it is.

Answer: A

The correct free body diagram of the lever is (A), as the key states.

The lever RS carries the applied force \(F\) downward at R. It rests against the left drum at T and against the right drum at U, and its foot S stands on the ground. The drums are smooth and the lever-drum friction is negligible, so each drum can push on the lever only perpendicular to it. The lever lies on the upper-right side of the left drum's centre, so at T the drum pushes the lever up and to the right (\(N_{1}\), perpendicular to the lever). The lever lies on the lower-left side of the right drum's centre, so at U that drum pushes the lever down and to the left (\(N_{2}\), also perpendicular). At S the ground gives a vertical reaction \(N_{3}\) and a horizontal friction force \(F_{f}\), because the problem makes only the drum-lever friction negligible and the unequal pushes at T and U have a horizontal component to be balanced.

(B) draws \(N_{1}\) vertical, but a smooth contact force must be perpendicular to the lever. (C) applies \(N_{2}\) at the foot S instead of at the contact point U. (D) leaves out the friction force at S. Only (A) has \(N_{1}\) at T and \(N_{2}\) at U perpendicular to the lever, with both \(N_{3}\) and \(F_{f}\) at S. The choice among the diagrams follows the printed arrows; I read them on the page image and did not find a published solution to compare against.