Question 1 |

A block of negligible mass rests on a surface that is inclined at 30^{\circ} to the horizontal plane as shown in the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just about to slide.

The coefficient of static friction between the block and surface is _____ (round off to two decimal places).

The coefficient of static friction between the block and surface is _____ (round off to two decimal places).

0.28 | |

0.36 | |

0.17 | |

0.05 |

Question 1 Explanation:

After forces are applied block is just about move (mass is negligible). Calculate coefficient of friction

\begin{aligned} F_{H} &=750 \mathrm{~N} \\ F_{V} &=900 \mathrm{~N} \\ \theta &=30^{\circ} \end{aligned}

\begin{aligned} N &=900 \cos \theta+750 \sin \theta \\ N &=900 \cos 30^{\circ}+750 \sin 30^{\circ} \\ &=1154.4228 \mathrm{~N} \\ F_{\max }+900 \sin 30^{\circ} &=750 \cos 30^{\circ} \\ \mu N &=199.519 \\ \mu &=\frac{199.519}{1154.4228} \\ \mu &=0.1728 \end{aligned}

Question 2 |

A plane truss PQRS(PQ=RS, \text{and }\angle PQR=90^{\circ}) is shown in the figure

The forces in the members PR andRS, respectively, are

The forces in the members PR andRS, respectively, are

F\sqrt{2} (tensile) and F (tensile) | |

F\sqrt{2} (tensile) and F (compressive) | |

F (compressive) and F\sqrt{2} (compressive) | |

F (tensile) and F\sqrt{2} (tensile) |

Question 2 Explanation:

Joint C,

\Sigma F_{H}=0

\begin{aligned} \Rightarrow \qquad F_{P R} \sin 45^{\circ}&=F \\ F_{P R}&=\sqrt{2} F(\text { Tensile }) \\ \Rightarrow\qquad F_{P R} \cos 45^{\circ}&=F_{R S}\\ \end{aligned}

\Sigma F_{V}=0

\Rightarrow F_{RS}=F(Comp.)

Question 3 |

A circular disk of radius r is confined to roll without slipping at P and Q as shown in
the figure.

If the plates have velocities as shown, the magnitude of the angular velocity of the disk is

If the plates have velocities as shown, the magnitude of the angular velocity of the disk is

\frac{v}{r} | |

\frac{v}{2r} | |

\frac{2v}{3r} | |

\frac{3v}{2r} |

Question 3 Explanation:

For pure rolling

\begin{array}{l} v_{P}=v=(P R) \omega \quad \ldots(i)\\ v_{Q}=2 v=(Q R) \omega \quad \ldots(ii) \end{array}

Divide by (ii) to (i),

2=\frac{Q R}{P R} \Rightarrow Q R=2(P R)

\begin{aligned} P R+Q R &=2 r \\ P R+2(P R) &=2 r \\ P R &=\frac{2}{3} r \\ \text{From equation(i) }\quad v &=\left(\frac{2}{3} r\right) \omega \Rightarrow \omega=\frac{3 v}{2 r} \end{aligned}

Question 4 |

An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal
force F as shown in the figure.

The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

A | |

B | |

C | |

D |

Question 4 Explanation:

Weigh = W

Note:

(i) When the cylinder is about to make out of the curb, it will loose its contact at point A, only contact will be at it B.

(ii) At verge of moving out of curb, Roller will be in equation under W, F and contact force from B and these three forces has to be concurrent so contact force from B will pass through C.

(iii) Even the surfaces are rough but there will be no friction at B for the said condition.

FBD

Question 5 |

The members carrying zero force (i.e. zero-force members) in the truss shown in the
figure, for any load P \gt 0 with no appreciable deformation of the truss (i.e. with no
appreciable change in angles between the members), are

BF and DH only | |

BF, DH and GC only | |

BF, DH, GC, CD and DE only | |

BF, DH, GC, FG and GH only |

Question 5 Explanation:

If at any joint three forces are acting out of which
two of them are collinear then force in third member
must be zero.

For member ED look at joint E.

Similarity look for other members.

For member ED look at joint E.

Similarity look for other members.

Question 6 |

A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value(s) of m is/are

1 kg only | |

6 kg only | |

1 kg, 6 kg | |

1 kg, 9 kg |

Question 6 Explanation:

Let V_{1} is the speed of 3 kg mass after collision

\mathrm{V}_{2} is the speed of m kg mass after collision

\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}

By linear momentum conservation

\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}

By using (1), (2) \& (3) we get

m = 1 kg (or) 9 kg

\mathrm{V}_{2} is the speed of m kg mass after collision

\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}

By linear momentum conservation

\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}

By using (1), (2) \& (3) we get

m = 1 kg (or) 9 kg

Question 7 |

The figure shows an idealized plane truss. If a horizontal force of 300 N is applied atpoint A, then the magnitude of the force produced in member CD is ______ N.

0 | |

10 | |

100 | |

200 |

Question 7 Explanation:

Adopting method of joints and taking FBD of joint B

F_{BC} = 0 (zero force member)

Further by taking FBD of joint C

F_{CD} = 0

Question 8 |

A truss is composed of members AB, BC, CD, AD and BD, as shown in the figure. A vertical load of 10 kN is applied at point D. The magnitude of force (in kN) in the member BC is____

10 | |

5 | |

15 | |

20 |

Question 8 Explanation:

BD is a zero force member

F.B.D of joint C is shown below,

For equilibrium of joint C

\begin{array}{l} \sum \mathrm{F}_{\mathrm{y}}=0, \quad \mathrm{F}_{\mathrm{CD}} \cos 45=5 \\ \quad \therefore \mathrm{F}_{\mathrm{CD}}=\frac{5}{\cos 45} \\ \sum \mathrm{F}_{\mathrm{x}=0}, \quad \mathrm{F}_{\mathrm{BC}}=\mathrm{F}_{\mathrm{CD}} \cos 45=\frac{5}{\cos 45} \times \cos 45=5 \mathrm{kN} \end{array}

F.B.D of joint C is shown below,

For equilibrium of joint C

\begin{array}{l} \sum \mathrm{F}_{\mathrm{y}}=0, \quad \mathrm{F}_{\mathrm{CD}} \cos 45=5 \\ \quad \therefore \mathrm{F}_{\mathrm{CD}}=\frac{5}{\cos 45} \\ \sum \mathrm{F}_{\mathrm{x}=0}, \quad \mathrm{F}_{\mathrm{BC}}=\mathrm{F}_{\mathrm{CD}} \cos 45=\frac{5}{\cos 45} \times \cos 45=5 \mathrm{kN} \end{array}

Question 9 |

A car having weight W is moving in the direction as shown in the figure. The center of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels is l. The acceleration of the car is a, and acceleration due to gravity is g. The reactions on the front wheels (R_f) and rear wheels (R_r) are given by

R_f=R_r=\frac{W}{2}-\frac{W}{g}\left ( \frac{h}{l} \right )a | |

R_f=\frac{W}{2}+\frac{W}{g}\left ( \frac{h}{l} \right )a; R_r=\frac{W}{2}-\frac{W}{g}\left ( \frac{h}{l} \right )a
| |

R_f=\frac{W}{2}-\frac{W}{g}\left ( \frac{h}{l} \right )a; R_r=\frac{W}{2}+\frac{W}{g}\left ( \frac{h}{l} \right )a
| |

R_f=R_r=\frac{W}{2}+\frac{W}{g}\left ( \frac{h}{l} \right )a |

Question 9 Explanation:

We analyse this problem in the frame of reference of car.

F.B.D of car is shown below,

As our frame of reference is accelerated hence, we have to apply a pseudo force 'ma' as shown

above, where, f_{1} and f_{2} are friction forces on rear and front wheels respectively.

For vertical equilibrium,

\begin{array}{l} \mathrm{R}_{\mathrm{r}}+\mathrm{R}_{\mathrm{f}}=\mathrm{W} \ldots(i)\\ \Sigma \mathrm{M}_{0}=0\\ W \times \frac{\ell}{2}-\frac{W a}{g} \times h-R_{f} \times \ell=0 \ldots(ii)\\ From (i) \& (ii)\\ R_{f}=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{\ell}\right) a \\ R_{r}=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{\ell}\right) a \end{array}

F.B.D of car is shown below,

As our frame of reference is accelerated hence, we have to apply a pseudo force 'ma' as shown

above, where, f_{1} and f_{2} are friction forces on rear and front wheels respectively.

For vertical equilibrium,

\begin{array}{l} \mathrm{R}_{\mathrm{r}}+\mathrm{R}_{\mathrm{f}}=\mathrm{W} \ldots(i)\\ \Sigma \mathrm{M}_{0}=0\\ W \times \frac{\ell}{2}-\frac{W a}{g} \times h-R_{f} \times \ell=0 \ldots(ii)\\ From (i) \& (ii)\\ R_{f}=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{\ell}\right) a \\ R_{r}=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{\ell}\right) a \end{array}

Question 10 |

A block of mass 10 kg rests on a horizontal floor. The acceleration due to gravity is 9.81 m/s^2. The coefficient of static friction between the floor and the block is 0.2. A horizontal force of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is___

5 | |

10 | |

12 | |

18 |

Question 10 Explanation:

Maximum friction force, f_{\max }=\mu \mathrm{N}=0.2 \times 10 \times 9.81=19.62 \mathrm{N}

Applied force, P=10 \mathrm{N} \lt \mathrm{f}_{\max }

\therefore Friction force = Applied force =10 \mathrm{N}

Applied force, P=10 \mathrm{N} \lt \mathrm{f}_{\max }

\therefore Friction force = Applied force =10 \mathrm{N}

There are 10 questions to complete.