Question 1 |

A cylindrical disc of mass m=1 \; kg and radius r=0.15\;m was spinning at \omega =5 \; rad/s when it was
placed on a flat horizontal surface and released
(refer to the figure). Gravity g acts vertically
downwards as shown in the figure. The coefficient
of friction between the disc and the surface is finite
and positive. Disregarding any other dissipation
except that due to friction between the disc and the
surface, the horizontal velocity of the center of the
disc, when it starts rolling without slipping, will be
_________ m/s (round off to 2 decimal places).

0 | |

0.25 | |

0.5 | |

0.75 |

Question 1 Explanation:

About point P there is no external torque.

[Because Torques of N and 'mg' balances and friction force pass through point P, hence its torque is zero]

\therefore \left [ \frac{d\vec{L}}{dt} =\vec{\tau }_{ext}\right ]_{\text{about P}}=0

where \vec{\tau }= Torque and \vec{L }= Angular momentum

\begin{aligned} \vec{L_i}&=\vec{L_f}\\ I_0\omega &=I_0\omega '+mvr\\ \frac{mr^2}{2}\omega &=\frac{mr^2}{2}\omega'+m\omega 'r^2\\ \therefore \omega '&=\frac{\omega }{3}=\frac{5}{3}\\ \Rightarrow v&=\omega ' \times r\\&=\frac{5}{3} \times 0.15=0.25 m/s \end{aligned}

Question 2 |

Two rigid massless rods PR and RQ are joined at
frictionless pin-joint R and are resting on ground
at P and Q, respectively, as shown in the figure. A
vertical force F acts on the pin R as shown. When
the included angle \theta \lt 90^{\circ} , the rods remain in static
equilibrium due to Coulomb friction between the
rods and ground at locations P and Q. At \theta = 90^{\circ} ,
impending slip occurs simultaneously at points P
and Q. Then the ratio of the coefficient of friction
at Q to that at P P(\mu _Q/\mu _P) is _________ (round off to
two decimal places).

5.76 | |

9.32 | |

4.25 | |

8.65 |

Question 2 Explanation:

By pythagoras theorem:

PQ = 13 m

By Geometry:

\begin{aligned} 5^2-PR^{'2}&=10^2-AR^{'2}\\ QR'&=13-PR'\\ \Rightarrow PR'&=\frac{25}{13}m\\ \Rightarrow QR'&=\frac{144}{13}m\\ \Sigma M_P&=0\\ \Rightarrow F(PR')&=N_Q(PQ)\\ \Rightarrow N_Q&=\frac{25}{169}F\\ \Sigma F&=0\\ \Rightarrow N_P&=F-\frac{25}{169}F\\ &=\frac{144}{169}F \end{aligned}

Now, f_P=\mu _PN_P, f_Q=\mu _QN_Q, \Sigma F_x=0

\begin{aligned} \mu _PN_P&=\mu _QN_Q\\ \frac{\mu _Q}{\mu _P}&=\frac{N_P}{N_Q}\\ &=\frac{144}{25}=5.76 \end{aligned}

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 6 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 7 |

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 7 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 8 |

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 8 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 9 |

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 9 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 10 |

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 10 Explanation:

\mu =0.1,m=1kg,F=0.8N

Now, from FBD as shown below figure

Normal reaction, N=mg=1 \times 9.81=9.81N

Limiting friction force between the block and the surface, f=\mu N=0.1 \times 9.81=0.98N

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

Now, from FBD as shown below figure

Normal reaction, N=mg=1 \times 9.81=9.81N

Limiting friction force between the block and the surface, f=\mu N=0.1 \times 9.81=0.98N

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

There are 10 questions to complete.

q10. correct option is a not b.

Thank You kkr,

We have updated the answer.

Correct option is B because block weight is more than applied force

Question Number?

Question no 10.

Correct option is B = 0.8 N (Body is in equilibrium)

How can friction be Zero , With Zero friction body will move on applying force

We have Update the answer.

14 question mistake