## A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h

Q: A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h(<l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30° with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/l and the frictional force f at the bottom of the stick are (g = 10 ms-2)

(A) $\frac{h}{l} = \frac{\sqrt{3}}{16} \; , f = \frac{16\sqrt{3}}{3}$

(B) $\frac{h}{l} = \frac{3}{16} \; , f = \frac{16\sqrt{3}}{3}$

(C) $\frac{h}{l} = \frac{3\sqrt{3}}{16} \; , f = \frac{8\sqrt{3}}{3}$

(D) $\frac{h}{l} = \frac{3\sqrt{3}}{16} \; , f = \frac{16\sqrt{3}}{3}$

Ans: D

Solution : $\displaystyle \Sigma F_x = 0$

$\displaystyle R_1 cos30^o – f = 0$

$\displaystyle R_1 cos30^o = f$ …(i)

$\displaystyle \Sigma F_y = 0$

$\displaystyle R_1 sin30^o + R_2 – m g = 0$

$\displaystyle R_1 sin30^o + R_2 = m g$

$\displaystyle \frac{R_1}{2} + R_1 = m g$ (Since R1 = R2)

$\displaystyle \frac{3 R_1}{2} = m g$

$\displaystyle R_1 = \frac{2}{3}m g = R_2$

$\displaystyle R_1 = \frac{2}{3}\times 16 = R_2$

$\displaystyle R_1 = \frac{32}{3} = R_2$

Hence , $\displaystyle f = R_1 cos30 = \frac{2}{3}m g \times \frac{\sqrt{3}}{2}$

$\displaystyle f = \frac{2}{3}\times 16 \times \frac{\sqrt{3}}{2}$

$\displaystyle f = \frac{16\sqrt{3}}{3}$

$\displaystyle \Sigma \tau_o = 0$

$\displaystyle mg \frac{l}{2}cos60^o – R_1(l – x) = 0$

$\displaystyle mg \frac{l}{2}\times \frac{l}{2} = \frac{2}{3}m g (l – x)$

$\displaystyle \frac{l}{4} = \frac{2}{3} (l – x)$

x = 5l/8

$\displaystyle cos30 = \frac{h}{l-x}$

$\displaystyle \frac{\sqrt{3}}{2} = \frac{h}{l-\frac{5l}{8}}$

$\displaystyle \frac{\sqrt{3}}{2} = \frac{h}{\frac{3l}{8}}$

$\displaystyle \frac{h}{l} = \frac{3\sqrt{3}}{16}$

## A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30° with horizontal) with constant velocity ….

Q: A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30° with horizontal) with constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by the inclined plane. ( take g = 10 m/s2 )

(A) 25 N

(B) 20 N

(C) 30 N

(D) none of these

Ans: (A)

Sol: Since the block slides down the incline with uniform
velocity, net force on it must be zero. Hence mg sinθ
must balance the frictional force ‘ f ‘ on the block.
Therefore f = mg sinθ = 5 × 10 × (1/2) = 25 N

## A body of mass 10 kg lies on a rough inclined plane of inclination θ = sin-(3/5) with the horizontal….

Q: A body of mass 10 kg lies on a rough inclined plane of inclination θ = sin(3/5) with the horizontal. When a force of 30 N is applied on the block parallel to & upward the plane, the total reaction by the plane on the block is nearly along

(A) OA

(B) OB

(C) OC

(D) OD

Ans: (A)

Solution: Frictional force along the in upward direction = 10 g sinθ – 30 = 30 N
N = 10 g cosθ = 80 NDirection of R is along OA

## A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between…

Q: A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between A & B is 0.5. (t is time in seconds. Initial velocities are zero, A is always on B)

(A) block A starts sliding on B at t = 5 seconds

(B) the heat produced due to friction in first 5 seconds is 312.5 J

(C) the heat produced due to friction in first 5 seconds is (625/8) J

(D) acceleration of A at 10 seconds is 5 m/s

Ans: (A) , (D)

Solution:

fmax = μ × 3 g = 0.5 × 30 = 15 N

block A starts sliding when friction force becomes max. i.e. fmax = 15 at that instant (F.B. D.)

both will move with same acceleration
So , 15 = 3a

a = 5 m/s2
F – 15 = 7 a

10 t – 15 = 7 × 5

10 t = 50

t = 5 sec
Work done by friction in 5 seconds

$W = \int F.dS = \int 10 t .dS$

$W = \int_{0}^{5} 10 t .v dt$ ;Since $v = \int a dt = \int t dt$ & a = F/m = 10t/10 = t ;

$W = \int_{0}^{5} 10 t .\frac{t^2}{2} dt$

$W = \int_{0}^{5} 5 t^3 dt$

$W = 5[\frac{t^4}{4}]_{0}^{5}$

$W = \frac{625 \times 5}{4} J$

## In the shown arrangement if f1, f2 and T be the frictional forces on 2 kg block, 3 kg block and tension in the string …

Q: In the shown arrangement if f1, f2 and T be the frictional forces on 2 kg block, 3 kg block and tension in the string respectively, then their values are:

(A) 2 N , 6 N , 3.2 N

(B) 2 N , 6 N, 0 N

(C) 1 N , 6 N , 2 N

(D) data insufficient to calculate the required values

Ans: (C)

Solution:

Net force without friction on system is 7 N in right side so first maximum friction will come on 3 kg block.

So f2 = 1 N, f3 = 6 N, T = 2

## Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm ….

Q: Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm . When you attempt to move both the fingers slowly towards the center , initially only the left finger slips with respect to the scale and right finger does not . After some distance , the left finger stops and right one starts slipping . Then the right finger stops at a distance xR from the center (50.00 cm) of the scale and the left one starts slipping again . This happen because of difference in  frictional forces on the two fingers . If the coefficient of static and dynamic friction between the fingers and scale are 0.40 and 0.32 respectively the value of xR (in cm ) is

Ans: (25.60)

Solution: μs = 0.40 , μk = 0.32

μs N4 = μkN3 ..(i)

x1 N3 = 40 × N4 …(ii)

On dividing ,

So , $\displaystyle \frac{\mu_k}{x_1} = \frac{\mu_s}{40}$

$\displaystyle \frac{0.32}{x_1} = \frac{0.40}{40}$

x1 = 32 cm

$\displaystyle \frac{\mu_s N_5}{x_1 N_5} = \frac{\mu_k N_6}{x_R N_6}$

xR = 25.60

## How does a lubricant help in reducing friction ?

Question: How does a lubricant help in reducing friction ?

Explanation : When a lubricant is added to a machine , it spreads between the two surfaces rubbing each other . On spreading of lubricant , the irregularities present on the surfaces are filled and it forms a thin layer between the surfaces in contact . As a result of it , the contact between the two hard surfaces is replaced by the contact between the hard surfaces and lubricant layer . Due to it the force of friction is reduced considerably .

## Blocks A and B of masses m1 and m2 are rest on the fixed wedge. The coefficient of friction between the wedge and the block is μ = 0.5 . If there is no motion then

Q: Blocks A and B of masses m1 and m2 are rest on the fixed wedge. The coefficient of friction between the wedge and the block is μ = 0.5 . If there is no motion then

(a) $\frac{2 – \sqrt{3}}{1 + 2 \sqrt{3}} \le \frac{m_1}{m_2} \le \frac{2 + \sqrt{3}}{2 \sqrt{3}-1}$

(b) $\frac{\sqrt{3}-1}{1 + 2 \sqrt{3}} \le \frac{m_1}{m_2} \le \frac{2 + \sqrt{3}}{2 \sqrt{3}-1}$

(c) $\frac{m_1}{m_2} \le \sqrt{3}$

(d) $\frac{m_1}{m_2} \ge \sqrt{3}$

Ans: (a)

## A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude…

Q: A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed a) 5 m/s b) 15 m/s Assume that friction is sufficient to prevent slipping (g = 10m/s2)

Click to See Solution :
Sol: v = 10 m/s

$\large tan\theta = \frac{v^2}{r g}$

$\large tan\theta = \frac{(10)^2}{20 \times 10 } = \frac{1}{2}$

Now, as speed is decreased, force of friction f acts upwards.

$\large N sin\theta – f cos\theta = \frac{m v^2}{r}$ ;

$\large N cos\theta + f sin\theta = m g$ ;

Substituting $\large tan\theta = \frac{1}{2}$ , v = 5 m/s, m = 200 kg and r = 20 m,

We get , f = 300√5 N

## A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between…

Q: A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.2. What is the maximum speed of the car for safe driving? (g = 10 ms-2)

Click to See Solution :
Sol: Maximum speed is

$\large v = \sqrt{\mu_s g r}$

$\large v = \sqrt{0.2 \times 10 \times 18}$

$\large v = \sqrt{36} = 6 m/s$