# Engineering Mechanics

 Question 1
A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction $(\mu)$ along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is $30^{\circ}$.

The limiting force P, in terms of Q, required for impending motion of block N to just move it in the upward direction is given as $P=\alpha Q$. The value of the coefficient '?' (rounded off to one decimal place) is
 A 0.6 B 0.5 C $2.0$ D 0.9
GATE CE 2021 SET-1
Question 1 Explanation:

\begin{aligned} \Sigma Y&=0 \Rightarrow \\ N_{2} \sin 60^{\circ}-0.2 N_{2} \sin &30^{\circ}-Q=0 \\ Q&=0.766 \mathrm{~N}_{2} \end{aligned}

\begin{aligned} \Sigma X &=0 \\ \Rightarrow \qquad 0.2 N_{2} \cos 30^{\circ}+N_{2} \cos 60^{\circ}-P &=0\\ P &=0.67 N_{2} \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P &=0.67 \times \frac{Q}{0.766}\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P&=0.875 \mathrm{Q} \simeq 0.9 Q \\ P&=\alpha Q \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \alpha&=0.9 \end{aligned}
 Question 2
A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacementof the system at any time t, the equation of motion for the free vibration of the system is $m\ddot{z}+kz=0$. The natural frequency of the system is
 A $\frac{k}{m}$ B $\sqrt{\frac{k}{m}}$ C $\frac{m}{k}$ D $\sqrt{\frac{m}{k}}$
GATE CE 2019 SET-1
Question 2 Explanation:
\begin{aligned} m\ddot{z}+kz&=0 \\ \ddot{z}+\frac{k}{m}z&=0 \\ \text{Comparing with} \\ \ddot{z}+\omega _n^2 z&=0 \\ \text{We get}\;\; \omega _n&=\sqrt{\frac{k}{m}} \end{aligned}
 Question 3
A cable PQ of length 25 m is supported at two ends at the same level as shown in the figure. The horizontal distance between the supports is 20 m. A point load of 150 kN is applied at point R which divides it into two equal parts.

Neglecting the self-weight of the cable, the tension (in kN, in integer value) in the cable due to the applied load will be _____
 A 100 B 125 C 200 D 250
GATE CE 2018 SET-2
Question 3 Explanation:

\begin{aligned} 12.5^{2} &=x^{2}+10^{2} \\ \Rightarrow \; \; x &=7.5\; \text{m} \end{aligned}
Bending moment at S=0 {consider the left part}
\begin{aligned} H_{P}\times 7.5 &=75\times 10 \\ H_{P} &=100\; \text{kN} \end{aligned}
Tension in cable
\begin{aligned} &=\sqrt{H_{P}^{2}+V_{P}^{2}}=\sqrt{75^{2}+100^{2}} \\ &=125\; \text{kN} \end{aligned}
 Question 4
Two rigid bodies of mass 5 kg and 4 kg are at rest on a frictionless surface until acted upon by a force of 36 N as shown in the figure. The contact force generated between the two bodies is
 A 4.0 N B 7.2 N C 9.0 N D 16.0 N
GATE CE 2018 SET-2
Question 4 Explanation:

As after action of 36 N, both blocks will move with same acceleration so considering 5 kg and 4 kg togather in a system and applying Newton's 2nd law.

$a=\frac{36}{9}=4\; m/s^{2}$

\begin{aligned} N &=ma\; \; \; \; \; \text{(Newton's 2nd law)} \\ N&=4\times 4=16\; \text{N} \end{aligned}
 Question 5
An aircraft approaches the threshold of a runway strip at a speed of 200 km/h. The pilot decelerates the aircraft at a rate of 1.697 m/$s^{2}$and takes 18 s to exit the runway strip. If the deceleration after exiting the runway is 1 m/$s^{2}$, then the distance (in m, up to one decimal place) of the gate position from the location of exit on the runway is ______
 A 312.8 B 360.4 C 210.4 D 111.8
GATE CE 2018 SET-1
Question 5 Explanation:
Speed of aircraft,
\begin{aligned} u_{i} &=200\; \text{km/hr} \\ &=\frac{200\times 1000}{3600}=55.56\; \text{m/s} \end{aligned}
Deceleration of aircraft on the runway $=1.697 m/s^{2}$
Aircraft takes 18 sec to exit the runway strip.
Speed of the aircraft at the exit of the runway
\begin{aligned} u_{f} &=u_{i}+at \\ &=55.56-1.697\times 18 \\ &=25.014\; \text{m/s} \end{aligned}
After runway aircraft decelerate with $1 m/s^{2}$
Total distance travelled by the aircraft from the location of exit on the runway
\begin{aligned} U^{2} &=u^{2}+2aS \\ 0 &=25.014^{2}-2\times 1\times S \\ S &=312.8\; \text{m} \end{aligned}
 Question 6
A cylinder of radius 250 mm and weight, W = 10 kN is rolled up an obstacle of height 50 mm by applying a horizontal force P at its centre as shown in the figure.

All interfaces are assumed frictionless. The minimum value of P is
 A 4.5 kN B 5.0 kN C 6.0 kN D 7.5 kN
GATE CE 2018 SET-1
Question 6 Explanation:

Given: r=250 mm, W=10 kN
Note:
1. When the cylinder will be about to move out of the cylinder, it will loose its contact at point A, only contact will be at point B.
2. Considering equation of cylinder of that instant under P, W and $R_{B}$ (contact force at B).
\begin{aligned} \sum \vec{M}_{B} &=0 \\ P\times OC-W\times BC &=0 \end{aligned}
In $\Delta OCB$
\begin{aligned} OC^{2}+CB^{2} &=OB^{2} \\ CB &=\sqrt{250^{2}-200^{2}}=150 \\ P &=\frac{10\times 10^{3}\times 150}{200} \\ &=2.5\times 3\: \text{kN}=7.5\: \text{kN} \end{aligned}
 Question 7
An assembly made of a rigid arm A-B-C hinged at end A and supported by an elastic rope C-D at end C is shown in the figure. The members may be assumed to be weightless and the lengths of the respective members are as shown in the figure.

Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the rope is
 A $\frac{3P}{\sqrt{2}}$ B $\frac{P}{\sqrt{2}}$ C $\frac{3P}{8}$ D $\sqrt{2}P$
GATE CE 2016 SET-2
Question 7 Explanation:

\begin{aligned} \sum M_{A}&=0\\ \Rightarrow \; R_{D}\times 2 L-P\times L&=0\\ \Rightarrow R_{D}&=\frac{P}{2} \end{aligned}
At joint D:

\begin{aligned} \sum F_{y}&=0\\ \Rightarrow \; T \cos 45^{\circ}&=\frac{P}{2}\\ \therefore \; T&=\frac{P}{\sqrt{2}} \end{aligned}
 Question 8
The magnitudes of vectors P, Q and R are 100 kN, 250 kN and 150 kN, respectively as shown in the figure

The respective values of Magnitude(in KN) and the Direction (with the respect to the x-axis) of the resultant vector are
 A 290.9 and 96.0$^{\circ}$ B 368.1 and 94.7$^{\circ}$ C 330.4 and 118.9$^{\circ}$ D 400.1 and 113.5$^{\circ}$
GATE CE 2016 SET-1
Question 8 Explanation:

Resultant force in x-direction,
\begin{aligned} R_{x}&=P\cos 60^{\circ}-Q\sin 15^{\circ}-R\sin 15^{\circ} \\ &=100\cos 60^{\circ}-250\sin 15^{\circ}-150\sin 15^{\circ} \\ &=-159.59\, kN\left ( \rightarrow \right ) \\ &=-159.59\, kN\left ( \rightarrow \right ) \\ &=159.59\, kN\left ( \leftarrow \right ) \end{aligned}
Resultant force in y-direction,
\begin{aligned} R_{y}&=P\sin 60^{\circ}+Q\cos 15^{\circ}-R\sin 15^{\circ} \\ &=100\sin 60^{\circ}+250\cos 15^{\circ}-150\sin 15^{\circ} \\ &=289.26\, kN \left ( \uparrow \right ) \end{aligned}
Final resultant force,
\begin{aligned} R&=\sqrt{R^{2}_{x}+R^{2}_{y}}=330.4\, kN \\ \theta &=\tan ^{-1}\left ( \frac{159.59}{289.56} \right ) \\ \theta &=28.9^{\circ} \end{aligned}

Direction with respect to x-axis
\begin{aligned} &=\theta+90^{\circ} \\ &=28.9^{\circ}+90^{\circ} \\ &=118.9^{\circ} \end{aligned}
 Question 9
Polar moment of inertia ($I_{p}$), in $cm^{4}$, of a rectangular section having width, b=2 and depth, d=6 cm is _____
 A 22 B 44 C 20 D 40
GATE CE 2014 SET-2
Question 9 Explanation:
Polar moment of inertia,
\begin{aligned} I_p&= I_x+I_y\\ &=\frac{bd^3}{12}+\frac{db^3}{12}\\ &=\frac{bd}{12}(b^2+d^2) \\ &= \frac{2 \times 6}{12}(2^2+6^2)\\ &=40\; cm^4 \end{aligned}
 Question 10
A disc of radius r has a hole of radius r/2 cut-out as shown. The centroid of the remaining disc (shaded portion) at a radial distance from the centre "O" is
 A r/2 B r/3 C r/6 D r/8
GATE CE 2010
Question 10 Explanation:
The centroid of the shaped portion of the disc is given by,
$\; \; \; x=\frac{A_{1}x_{1}+A_{2}x_{2}}{A_{1}+A_{2}}$
where x is the radial distance from O.
\begin{aligned} A_{1}&=\, \pi r^{2};\; x_{1}=0;\\ A_{2}&=-\pi \times \left ( \frac{r}{2} \right )^{2}=-\frac{\pi r^{2}}{4}\\ x_{2}&=\frac{r}{2}\\ \therefore \; x&=\frac{\pi r^{2}\times 0-\frac{\pi r^{2}}{4}\times \frac{r}{2}}{\pi r^{2}-\frac{\pi r^{2}}{4}}\\ x&=\frac{\pi r^{2}\times 0-\frac{\pi r^{2}}{4}\times \frac{r}{2}}{\pi r^{2}-\frac{\pi r^{2}}{4}}\\ \Rightarrow \; x&=-\frac{r}{6} \end{aligned}
There are 10 questions to complete.