Your Answer : (Not Answered)

Correct Answer : D

The particle must satisfy both \(\Sigma f_{x} = 0\) and \(\Sigma f_{y} = 0\).

For a particle in equilibrium under coplanar concurrent forces, the vector sum of the forces must be zero, \(\Sigma \vec{F} = 0\). In the X-Y plane that one vector equation splits into two scalar equations, one along each axis.

If only \(\Sigma f_{x} = 0\) held, a resultant along Y would remain, and the particle would accelerate in that direction; the same is true the other way round. The first option, with \(\Sigma f_{y} \neq 0\), describes a particle that is not in equilibrium at all. No moment equation is needed, because concurrent forces all pass through one point.

Your Answer : (Not Answered)

Correct Answer : D

The second moment of area about a diameter is \(\frac{\pi D^{4}}{64}\).

The polar moment about the centre is \(J = \frac{\pi D^{4}}{32}\). By the perpendicular axis theorem it equals the sum of the moments about two perpendicular diameters, which are equal by symmetry.

\[ I_{d} = \frac{J}{2} = \frac{\pi D^{4}}{64} \]

\(\pi D^{4}/32\) is the polar moment, the usual trap. \(\pi D^{4}/4\) and \(\pi D^{4}/16\) do not correspond to the second moment of a circle about any of these axes.

Your Answer : (Not Answered)

Correct Answer : C

The moment of inertia is \(0.11\, r^4\).

About its diameter, a semicircle has \(I = \frac{\pi r^4}{8}\). Its centroid lies \(\bar{y} = \frac{4r}{3\pi}\) from the diameter, so the parallel axis theorem gives the centroidal value:

\[ \begin{aligned} I_{xx} &= \frac{\pi r^4}{8} - \frac{\pi r^2}{2}\left(\frac{4r}{3\pi}\right)^2 \\ &= 0.3927r^4 - 0.2829r^4 \\ &= 0.11\, r^4 \end{aligned} \]

0.22 r⁴ is twice the value, and 0.055 r⁴ half of it. Forgetting to subtract the parallel axis term would leave 0.39 r⁴, which is not offered.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : B

The correct free body diagram is (B).

A free body diagram shows the block alone, with every external force: W at the centre of gravity, P through the centre of gravity, friction F at the base opposing P, and the normal reaction N. Horizontal balance gives F = P and vertical balance N = W.

F acts at the base, below the line of P, so P and F form a couple that tends to tip the block forward. To balance it, N must act at a point shifted forward of W's line, in the direction of P, by \(x = Ph/W\) for P at height h. Diagram (B) shows this shift.

In (A) N and W are collinear, leaving the P-F couple unbalanced. In (C) N is shifted the wrong way. (D) draws the supporting surface, which a free body diagram must leave out, and has N in line with W.

Your Answer : (Not Answered)

Correct Answer : B

The minimum effort is \(W\left(\frac{\mu_{s} + \tan\theta}{1 - \mu_{s}\tan\theta}\right)\).

When upward motion impends, friction \(\mu_{s}N\) acts down the slope. Resolve along and normal to the incline for the horizontal force P:

\[ \begin{aligned} P\cos\theta &= W\sin\theta + \mu_{s}N \\ N &= W\cos\theta + P\sin\theta \end{aligned} \]

Substituting and dividing by \(\cos\theta\) gives \(P(1 - \mu_{s}\tan\theta) = W(\tan\theta + \mu_{s})\), the keyed expression (equal to \(W\tan(\theta + \phi)\) with \(\tan\phi = \mu_{s}\)).

The expressions with \(\mu_{s} - \tan\theta\) belong to a block about to slide down, and the one with \(1 + \mu_{s}\tan\theta\) in the denominator gets the sign of the normal-force contribution of P wrong.

Your Answer : (Not Answered)

Correct Answer : A

The block starts to slide when F = \(\mu W\).

Spring \(k_{2}\) transmits the full pull F to the block, whatever its stiffness, since it is massless and in series with the applied force. Spring \(k_{1}\) is tied to the wall and is undeformed at the start; it can only exert a force once the block moves and stretches it.

So while the block is at rest, static friction alone resists F, and slipping begins when F reaches the limiting friction \(\mu W\). The options with stiffness ratios would apply only if the block had already been displaced, so that \(k_{1}\) were carrying part of the load.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

Her angular velocity increases.

The ice exerts almost no torque about the vertical axis, so the skater's angular momentum \(L = I\omega\) is conserved. Pulling her arms and legs in brings mass closer to the axis and reduces her moment of inertia I, so \(\omega\) must increase to keep \(I\omega\) constant.

A decrease would happen if she extended her limbs. The angular velocity would stay the same only if I were unchanged, and it would go to zero only under an external braking torque.

Your Answer : (Not Answered)

Correct Answer : C

The non-rigid frame is (C).

A pin-jointed frame is rigid when it is built up from triangles, each new joint fixed by two bars to points already fixed. In frame (C) the left bay has a diagonal and the right bay is braced by crossing bars through a central pin, but the middle bay is a plain quadrilateral with no diagonal. Four bars pinned at their ends form a mechanism: the middle bay can sway into a parallelogram, carrying the right bay with it, so the frame as a whole is not rigid.

Frame (A) is two triangles sharing a bar. Frame (B) is a chain of triangles, each sharing an edge with the last, closed by a final triangle at the bottom right. Frame (D) has a diagonal in every bay. All three are rigid.

Your Answer : (Not Answered)

Correct Answer : C

The ram should be released from 35.1°.

Because the two ropes are equal and parallel, the ram does not rotate: it moves in curvilinear translation, every point travelling on a circle of radius 7 m. Its size (9 m by 2 m) therefore does not enter. When released from angle \(\theta\), its centre of mass falls by \(7(1 - \cos\theta)\), and energy conservation gives the speed at the bottom:

\[ \begin{aligned} v^{2} &= 2gL(1 - \cos\theta) \\ 1 - \cos\theta &= \frac{25}{2 \times 9.81 \times 7} = 0.1820 \\ \cos\theta &= 0.8180 \Rightarrow \theta = 35.1^{\circ} \end{aligned} \]

The larger angles would give striking speeds well above 5 m/s; for example 67.1° corresponds to about 9.2 m/s.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : B

The force must be \(\frac{mg}{2}\).

Both bars are horizontal with F acting downward at the free end, so gravity adds to F in the end-pinned case but has no moment about the centre pin in the second case.

\[ \begin{aligned} \alpha_1 &= \frac{2FL + mgL}{\tfrac{4}{3}mL^{2}} = \frac{3(2F + mg)}{4mL} \\ \alpha_2 &= \frac{FL}{\tfrac{1}{3}mL^{2}} = \frac{3F}{mL} \end{aligned} \]

using \(I = m(2L)^{2}/3\) about an end and \(I = m(2L)^{2}/12\) about the centre. Setting \(\alpha_1 = \alpha_2\) gives \(2F + mg = 4F\), so \(F = mg/2\).

Ignoring gravity would make the two accelerations unequal for any F, and mg, mg/4 and mg/6 do not satisfy the balance above.

Your Answer : (Not Answered)

Correct Answer : D

The speed of the centre after impact is \(0.9v\).

The impulse at S passes through S, and there is no impulse from PQ, so angular momentum about S is conserved during the impact. Before impact the centre moves horizontally at \(v\), at height \(R - \frac{3R}{20} = \frac{17R}{20}\) above S, and the disk spins at \(\omega = v/R\):

\[ \begin{aligned} H_{S} &= mv\tfrac{17R}{20} + \tfrac{1}{2}mR^{2}\tfrac{v}{R} \\ &= \tfrac{27}{20}mvR \end{aligned} \]

After impact the disk pivots about S without slip or rebound, with moment of inertia about S of \(\frac{1}{2}mR^{2} + mR^{2} = \frac{3}{2}mR^{2}\):

\[ \begin{aligned} \tfrac{3}{2}mR^{2}\omega' &= \tfrac{27}{20}mvR \\ \omega' &= 0.9\tfrac{v}{R}, \quad v' = \omega' R = 0.9v \end{aligned} \]

Leaving out the spin term \(\frac{1}{2}mR^{2}\omega\) would give only about \(0.57v\); the disk's own rotation carries angular momentum into the impact, which is why the speed falls by only a tenth. 0.1v, 0.3v and 0.7v would all require angular momentum about S to be lost, which the stated conditions rule out.

Your Answer : (Not Answered)

Correct Answer : B

The angular acceleration is \(\frac{F}{2mc}\).

On a smooth floor the assembly is free. Its centre of mass is at O, and the moment of inertia about O is that of four masses at distance c:

\[ \begin{aligned} I_{O} &= 4mc^{2} \\ \tau_{O} &= 2F \cdot c \\ \alpha &= \frac{2Fc}{4mc^{2}} = \frac{F}{2mc} \end{aligned} \]

The force also accelerates the centre of mass, at \(2F/4m\), but that does not change the rotation about the centre of mass. \(F/(mc)\) uses only two of the masses in the inertia, and \(2F/(3mc)\) and \(F/(3mc)\) do not match the inertia \(4mc^{2}\) of four equal masses.

Your Answer : (Not Answered)

Correct Answer : A

The velocity of A is \((1 - 2\sqrt{3})\)i + 2j m/s.

The rod makes 60° with the x-axis and is 2 m long, so the position of A relative to B is

\[ \begin{aligned} \mathbf{r}_{A/B} &= 2\cos 60^{\circ}\,\mathbf{i} + 2\sin 60^{\circ}\,\mathbf{j} \\ &= \mathbf{i} + \sqrt{3}\mathbf{j} \end{aligned} \]

It turns anticlockwise, \(\boldsymbol{\omega} = 2\mathbf{k}\):

\[ \begin{aligned} \mathbf{v}_{A} &= \mathbf{v}_{B} + \boldsymbol{\omega} \times \mathbf{r}_{A/B} \\ &= \mathbf{i} + 2\mathbf{k} \times (\mathbf{i} + \sqrt{3}\mathbf{j}) \\ &= \mathbf{i} + 2\mathbf{j} - 2\sqrt{3}\mathbf{i} \end{aligned} \]

\(-2\sqrt{3}\)i + 2j leaves out the velocity of B, and the options with \(-2\)j take the rotation as clockwise.

Your Answer : (Not Answered)

Correct 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.