# Friction

 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). A 0 B 0.25 C 0.5 D 0.75
GATE ME 2022 SET-1   Engineering Mechanics
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). A 5.76 B 9.32 C 4.25 D 8.65
GATE ME 2022 SET-1   Engineering Mechanics
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).
 A 0.28 B 0.36 C 0.17 D 0.05
GATE ME 2021 SET-2   Engineering Mechanics
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___ A 5 B 10 C 12 D 18
GATE ME 2019 SET-1   Engineering Mechanics
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 $=10 \mathrm{N}$
 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? A 490.5 and 0.5 B 981 and 0.5 C 1000.5 and 0.15 D 1000.5 and 0.25
GATE ME 2014 SET-4   Engineering Mechanics
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} 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 _______ A 57.664m B 98.215m C 25.154m D 32.459m
GATE ME 2014 SET-3   Engineering Mechanics
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 _______
 A 0.1948m/s B 2.154m/s C 3.2145m/s D 1.1948m/s
GATE ME 2014 SET-2   Engineering Mechanics
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 _______
 A 5.9m/s B 4.9m/s C 3.9m/s D 2.5m/s
GATE ME 2014 SET-1   Engineering Mechanics
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}
 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 A 0.69 B 0.88 C 0.98 D 1.37
GATE ME 2014 SET-1   Engineering Mechanics
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 A 0 B 0.8N C 0.98N D 1.2N
GATE ME 2011   Engineering Mechanics
Question 10 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 $0.8 \mathrm{N}$
There are 10 questions to complete.

### 2 thoughts on “Friction”

1. q10. correct option is a not b.

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