Engineering Mechanics Questions and Answers by Topic

Answer: D

The angle is \(\theta = \frac{\pi}{6}\).

The force F is normal to the sloping face, which makes angle \(\theta\) with the horizontal base, so F makes the same angle \(\theta\) with the vertical. Its horizontal component pushes the block towards the wall and is resisted by the spring; the vertical component is taken by the floor.

\[ \begin{aligned} F\sin\theta &= k\Delta = 100 \times 0.01 = 1\ \text{N} \\ \sin\theta &= \frac{1}{2},\ \theta = \frac{\pi}{6} \end{aligned} \]

Using the cosine instead of the sine gives \(\pi/3\), the complementary angle. \(\pi/4\) and \(\pi/5\) give horizontal components of 1.41 N and 1.18 N, which do not match the spring force of 1 N.

Answer: A

The Coriolis acceleration is 1.2 \(\text{m/s}^{2}\).

For a body moving with velocity \(v\) relative to a frame rotating at \(\omega\), with \(v\) perpendicular to the rotation axis, the Coriolis component is \(2\omega v\):

\[ \begin{aligned} a_{c} &= 2\omega v = 2 \times 3 \times 0.2 \\ &= 1.2\ \text{m/s}^{2} \end{aligned} \]

0.6 omits the factor 2, 2.4 doubles it again, and 0.3 is neither. The direction is perpendicular to the slot, but the question asks only for the magnitude.

Answer: C

The resultant makes an angle of 60° with the positive x-axis.

The two forces are perpendicular, F along x and \(\sqrt{3}F\) along y. Their resultant has components (F, \(\sqrt{3}F\)), so:

\[ \begin{aligned} \tan\theta &= \frac{\sqrt{3}F}{F} = \sqrt{3} \\ \theta &= 60^{\circ},\quad |R| = 2F \end{aligned} \]

30° is the angle the resultant makes with the y-axis, a slip from swapping the components. 45° would need equal components, and 75° has no basis.

Answer: A

The zero-force member is ST.

Look at joint T. It meets only two members, ST (horizontal) and RT (vertical), and the roller support there gives only a vertical reaction. There is no horizontal external force at T, so horizontal equilibrium of the joint gives the force in ST as zero. RT then carries the support reaction.

The rule of thumb: at an unloaded joint, or one loaded only along one member, where two non-collinear members meet, the member not aligned with the load carries nothing.

SR, SP and SQ all carry load: the 10 kN acts at S, and SQ, SP and SR are needed to carry it to the supports.

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.