Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 6 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 7 |

A ball is dropped from rest from a height of 1 m in a frictionless tube as shown in the figure. If the tube profile is approximated by two straight lines (ignoring the curved portion), the total distance travelled (in m) by the ball is __________ (correct to two decimal places).

Marks to all | |

Marks to all | |

Marks to all | |

Marks to all |

Question 7 Explanation:

\begin{aligned} \frac{B C}{A B} &=\sin 45^{\circ} \\ A B &=\frac{B C}{\sin 45^{\circ}}=\frac{1}{\sin 45^{\circ}}=1.4142 \mathrm{m} \\ \text { Total travel, } O A+A B &=1+1.4142 \mathrm{m}=2.414 \mathrm{m} \end{aligned}

Question 8 |

Block P of mass 2 kg slides down the surface and has a speed 20 m/s at the lowest point, Q, where the local radius of curvature is 2 m as shown in the figure. Assuming g=10 m/s^{2}, the normal force (in N) at Q is _______ (correct to two decimal places).

230 | |

350 | |

420 | |

533 |

Question 8 Explanation:

m = 2 kg,\; g = 10m/s^{2}

FBD of mass P at point Q

N-m g=m a_{r} \quad at no slip condition

N=m g+\frac{m v^{2}}{R}=2 \times 10+\frac{2 \times 20 \times 20}{2}=420 \mathrm{N}

Question 9 |

A bar of uniform cross section and weighing 100 N is held horizontally using two massless and inextensible strings S1 and S2 as shown in the figure.

The tension of the strings are

The tension of the strings are

T_{1} = 100 N \; and \; T_{2} = 0 N | |

T_{1} = 0 N \; and \; T_{2} = 100 N | |

T_{1} = 75 N \; and \; T_{2} = 25 N | |

T_{1} = 25 N \; and \; T_{2} = 75 N |

Question 9 Explanation:

\begin{aligned} T_{1}+T_{2} &=100 \mathrm{N} \qquad \ldots(i)\\ \Sigma M_{A} &=0 \\ T_{2} \cdot \frac{L}{2} &=100 \times \frac{L}{2} \\ \therefore \qquad T_{2} &=100 \mathrm{N} \\ T_{1} &=0 \mathrm{N} \end{aligned}

Question 10 |

A force F is acting on a bent bar which is clamped at one end as shown in the figure.

The CORRECT free body diagram is

The CORRECT free body diagram is

A | |

B | |

C | |

D |

Question 10 Explanation:

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.

Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.

Option (C) is wrong because x-component is not shown.

Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.

Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.

Option (C) is wrong because x-component is not shown.

Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.

There are 10 questions to complete.

Solution to question no. 10 is incorrect. Please check.

Thank You Gliesha,

We have updated the solution.