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 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 2 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}

Question 3 |

A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of friction \mu between the floor and the wardrobe, respectively?

490.5 and 0.5 | |

981 and 0.5 | |

1000.5 and 0.15 | |

1000.5 and 0.25 |

Question 3 Explanation:

We can solve this question by moment or by force

triangle as this is condition of equilibrium (just to topple)

\begin{aligned} \sum M_{Q} &=0 \\ P \times 2-m g \times 1 &=0 \\ P &=\frac{m g}{2}=\frac{981}{2} \\ &=490.5 \mathrm{N} \end{aligned}

as P is horizontal force and it will be over come

by friction force

P=\left(f_{s}\right)_{\max } [ as this is at verge of motion ]

\begin{aligned} 490.5 &=\mu_{s} m g \\ \mu_{s} &=0.5 \end{aligned}

\begin{aligned} \tan \theta &=\frac{2}{1} \\ \theta &=63.4349^{\circ} \end{aligned}

Wardrobe is about to topple about pt Q so there

will be contact force from point Q on the wardrobe

\begin{aligned} \frac{W}{P} &=\tan \theta=2 \\ P &=\frac{W}{2}=\frac{981}{2}=490.5 \mathrm{N} \end{aligned}

Two components of R will balance W and P and

horizontal component (parallel to contacting

surface is called friction and perpendicular to

contacting surface is called normal reaction)

\begin{aligned} R \cos \theta &=\text { friction force } \\ &=P(\text { in magnitude }) \\ \left(f_{s}\right)_{\max } &=P=490.5 \\ \text{and}\qquad \left(f_{s}\right)_{\max } &=\mu_{s} m g\\ \mu_{s} m g &=490.5 \\ \mu_{s} &=0.5 \end{aligned}

triangle as this is condition of equilibrium (just to topple)

\begin{aligned} \sum M_{Q} &=0 \\ P \times 2-m g \times 1 &=0 \\ P &=\frac{m g}{2}=\frac{981}{2} \\ &=490.5 \mathrm{N} \end{aligned}

as P is horizontal force and it will be over come

by friction force

P=\left(f_{s}\right)_{\max } [ as this is at verge of motion ]

\begin{aligned} 490.5 &=\mu_{s} m g \\ \mu_{s} &=0.5 \end{aligned}

**OR**\begin{aligned} \tan \theta &=\frac{2}{1} \\ \theta &=63.4349^{\circ} \end{aligned}

Wardrobe is about to topple about pt Q so there

will be contact force from point Q on the wardrobe

\begin{aligned} \frac{W}{P} &=\tan \theta=2 \\ P &=\frac{W}{2}=\frac{981}{2}=490.5 \mathrm{N} \end{aligned}

Two components of R will balance W and P and

horizontal component (parallel to contacting

surface is called friction and perpendicular to

contacting surface is called normal reaction)

\begin{aligned} R \cos \theta &=\text { friction force } \\ &=P(\text { in magnitude }) \\ \left(f_{s}\right)_{\max } &=P=490.5 \\ \text{and}\qquad \left(f_{s}\right)_{\max } &=\mu_{s} m g\\ \mu_{s} m g &=490.5 \\ \mu_{s} &=0.5 \end{aligned}

Question 4 |

A body of mass (M) 10 kg is initially stationary on a 45^{\circ} inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______

57.664m | |

98.215m | |

25.154m | |

32.459m |

Question 4 Explanation:

\begin{aligned} & v^{2}=u^{2}+2 a s \\ & u=0 \quad(\text { given }) \\ \therefore \quad & \quad s=\frac{v^{2}}{2 a}=\frac{v^{2}}{2 \mu g \cos 45^{\circ}} \\ \therefore \quad & \quad s=\frac{400}{2 \times 0.5 \times 9.81 \times \cos 45^{\circ}} \\ &=57.664 \mathrm{m} \end{aligned}

Question 5 |

A truck accelerates up a 10^{\circ} incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3. The maximum value of acceleration (in m/s^{2} ) of the truck such that the crate does not slide down is _______

0.1948m/s | |

2.154m/s | |

3.2145m/s | |

1.1948m/s |

Question 5 Explanation:

Crate will acquire max acceration when \left(f_{s}\right)_{\max } will

be acting on crate by truck and at that condition

acceleration of truck and crate wil be same.

Increasing acceleration of truck after that will cause

slipping of crate on truck.

\left[\left(f_{s}\right)_{\max }\right]_{C T}=\mu N_{C T}

\begin{aligned}\left[\left(f_{s}\right)_{\max }\right]_{C r}-m g \sin 10^{\circ} &=m a \\ \mu m g \cos 10^{\circ}-m g \sin 10^{\circ} &=m a \\ a &=1.1948 \mathrm{m} / \mathrm{s}^{2} \end{aligned}

Question 6 |

A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in newton) P=10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is _______

5.9m/s | |

4.9m/s | |

3.9m/s | |

2.5m/s |

Question 6 Explanation:

\begin{aligned} N &=m g=200 \mathrm{N} \\(f_{s})_{max} &=\mu_{s} N \\ &=0.4 \times 200=80 \end{aligned}

upto 8 \sec [P=10 t] \rightarrow no motion

\Rightarrow from 8 sec to 10 sec

\begin{aligned} f_{k} &=\mu_{k} N \\ &=0.2 \times 200=40 \mathrm{N} \end{aligned}

Average applied force

=\frac{P_{\max }+P_{\min }}{2}

=\frac{100+80}{2}=90 \mathrm{N}

Average resultant force = Avg. applied force kinetic friction

=90-40=50 N

Average acceration

\begin{aligned} &=\frac{50 \times 9.8}{200} \\ \Rightarrow \quad V &=u+a t\\ u&=0\\ a&=\frac{50 \times 9.8}{200} \\ t&=2 \text{second}\\ \therefore V&=\frac{50 \times 9.8}{200} \times 2=4.9 \mathrm{m} / \mathrm{s} \end{aligned}

upto 8 \sec [P=10 t] \rightarrow no motion

\Rightarrow from 8 sec to 10 sec

\begin{aligned} f_{k} &=\mu_{k} N \\ &=0.2 \times 200=40 \mathrm{N} \end{aligned}

Average applied force

=\frac{P_{\max }+P_{\min }}{2}

=\frac{100+80}{2}=90 \mathrm{N}

Average resultant force = Avg. applied force kinetic friction

=90-40=50 N

Average acceration

\begin{aligned} &=\frac{50 \times 9.8}{200} \\ \Rightarrow \quad V &=u+a t\\ u&=0\\ a&=\frac{50 \times 9.8}{200} \\ t&=2 \text{second}\\ \therefore V&=\frac{50 \times 9.8}{200} \times 2=4.9 \mathrm{m} / \mathrm{s} \end{aligned}

Question 7 |

A block R of mass 100kg is placed on a block S of mass 150kg as shown in the figure. Block R is tied to the wall by a massless and intextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F(in kN) needed to move the block S is

0.69 | |

0.88 | |

0.98 | |

1.37 |

Question 7 Explanation:

\begin{aligned} F &=\mu\left(W_{S}+W_{R}\right)+\mu W_{R} \\ F &=\mu\left(W_{S}+2 W_{R}\right) \\ &=0.4(150+2 \times 100) \times 9.81 \\ &=1373.4 \mathrm{N}=1.37 \mathrm{kN} \end{aligned}

Question 8 |

A 1 kg block is resting on a surface with coefficient of friction \mu =0.1
A force of 0.8N is applied to the block as shown in figure. The friction force is

0 | |

0.8N | |

0.98N | |

1.2N |

Question 8 Explanation:

\begin{aligned} \text { Friction force } &=\mu \times N \\ \Sigma F_{V} &=0\\ \therefore \qquad N-m g &=0 \\ \text{Friction force} &=m g=1 \times 9.8 \\ &=0.1 \times 9.8 \\ &=0.98 \mathrm{N} \end{aligned}

since applied force is less than friction force, hence friction force acting on body at equlibrium is 0.8 \mathrm{N}

Question 9 |

A block weighing 981N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is \mu=0.2A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?

176.2 | |

196 | |

481 | |

981 |

Question 9 Explanation:

Free body Diagram

Assume normal reaction =R (Newton)

Balancing force in horizontal direction

(\Sigma F x=0)

\begin{aligned} \mu R &=100 \\ (0.2) R &=100 \\ R &=500 \mathrm{N} \end{aligned}

Balancing the force in vertical direction

\begin{aligned} (\Sigma F y&=0) \\ T+R-W &=0 \\ \Rightarrow T+500-981 &=0 \\ T &=481 \mathrm{N} \end{aligned}

Assume normal reaction =R (Newton)

Balancing force in horizontal direction

(\Sigma F x=0)

\begin{aligned} \mu R &=100 \\ (0.2) R &=100 \\ R &=500 \mathrm{N} \end{aligned}

Balancing the force in vertical direction

\begin{aligned} (\Sigma F y&=0) \\ T+R-W &=0 \\ \Rightarrow T+500-981 &=0 \\ T &=481 \mathrm{N} \end{aligned}

Question 10 |

A block of mass M is released from point P on a rough inclined plane with inclination angle \theta, shown in the figure below. The co - efficient of friction is \mu . If \mu \lt \tan\theta, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is

\sqrt{\frac{2s}{g \cos\theta (\tan \theta - \mu ) }} | |

\sqrt{\frac{2s}{g \cos\theta (\tan \theta + \mu ) }} | |

\sqrt{\frac{2s}{g \sin\theta (\tan \theta - \mu ) }} | |

\sqrt{\frac{2s}{g \sin\theta (\tan \theta + \mu ) }} |

Question 10 Explanation:

FBD of block:

N=m g \cos \theta

From Newton's second law

mg \sin \theta-\mu m g \cos \theta=m a

a=g \sin \theta-\mu g \cos \theta

S=u t+\frac{1}{2} a t^{2}

S=0+\frac{1}{2}(g \sin \theta-\mu g \cos \theta) t^{2}

t=\sqrt{\frac{2 s}{g \cos \theta[\tan \theta-\mu]}}

N=m g \cos \theta

From Newton's second law

mg \sin \theta-\mu m g \cos \theta=m a

a=g \sin \theta-\mu g \cos \theta

S=u t+\frac{1}{2} a t^{2}

S=0+\frac{1}{2}(g \sin \theta-\mu g \cos \theta) t^{2}

t=\sqrt{\frac{2 s}{g \cos \theta[\tan \theta-\mu]}}

There are 10 questions to complete.

q10. correct option is a not b.

Thank You kkr,

We have updated the answer.