Engineering Mechanics Questions and Answers

Answer: B

Let \(\theta\) be the angle between the vectors and \(c = \cos\theta\). Then \(R^2 = A^2 + B^2 + 2ABc\). Doubling B gives \(4R^2 = A^2 + 4B^2 + 4ABc\), and reversing B gives \(4R^2 = A^2 + B^2 - 2ABc\). Subtracting the last two: \(3B^2 + 6ABc = 0\), so \(2ABc = -B^2\).

So \(A : B : R = 1 : \sqrt{3/2} : 1 = \sqrt{2} : \sqrt{3} : \sqrt{2}\). The ratios 2:3:2 and 2:1:3 are not consistent with R equal to A.

Answer: A

Coplanar forces are forces whose lines of action all lie in one plane, which is the first option. Forces acting at a single point are concurrent forces, a single force that keeps the body in equilibrium together with other forces is the equilibrant, and a single force that produces the same effect as two or more forces is the resultant. So the other three options define different terms. In coplanar force systems all the forces can be shown on a flat drawing, and this is the common case in statics problems.

Answer: A

All three statements are true. Physical quantities with only magnitude, such as mass and time, are scalars. Quantities with both magnitude and direction, such as velocity and force, are vectors. Two equal and opposite parallel forces acting at different points of a body constitute a couple, which tends to turn the body without moving it as a whole. Since all three are true, the options that mark any of them false are wrong. The turning effect of a couple is its moment, which is the force times the perpendicular distance between the two lines of action.

Answer: B

The third load must act in the x - y plane but not along x or y axis.

The load at A acts along the x-axis and A lies on that axis, so its line of action passes through O. Likewise the load at B acts along the y-axis, through O. The third load acts at O itself. All three lines pass through O, so the moments about O are zero automatically, and equilibrium needs only the forces to sum to zero:

\[ \mathbf{F}_{O} = -(F_{A}\,\hat{i} + F_{B}\,\hat{j}) \]

With both given loads non-zero, this force has x and y components and no z component: it lies in the x-y plane but along neither axis. A force along the z-axis, or in the y-z or z-x planes, would leave an unbalanced component in x or y.

Answer: B

The equivalent system is \(\vec{F}_{O} = (2\vec{i} + \vec{j})N\) and \(\vec{M}_{O} = 0\).

Moving a force to the origin leaves the force unchanged and adds its moment about the origin. A couple is a free vector, so it moves unchanged wherever it acts.

\[ \begin{aligned} \vec{r}_{1} \times \vec{F} &= (\vec{i} + 2\vec{j}) \times (2\vec{i} + \vec{j}) \\ &= (1 \cdot 1 - 2 \cdot 2)\vec{k} = -3\vec{k} \\ \vec{M}_{O} &= -3\vec{k} + 3\vec{k} = 0 \end{aligned} \]

The option with \(3\vec{k}\) forgets the moment of the force, the one with \(6\vec{k}\) gets its sign wrong, and reversing the force to \(-2\vec{i} - \vec{j}\) is never right, because an equivalent system keeps the same resultant.

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.