Advertisements
Advertisements
प्रश्न
Consider the situation shown in the following figure. Both the pulleys and the string are light and all the surfaces are frictionless.
- Find the acceleration of the mass M.
- Find the tension in the string.
- Calculate the force exerted by the clamp on the pulley A in the figure.

Advertisements
उत्तर
Let the acceleration of mass M be a.
So, the acceleration of mass 2M will be \[\frac{a}{2}\]

(a)

2M(a/2) − 2T = 0
⇒ Ma = 2T
T + Ma − Mg = 0
\[\Rightarrow \frac{Ma}{2} + Ma = Mg \]
\[ \Rightarrow 3Ma = 2Mg\]
\[ \Rightarrow a = \frac{2g}{3}\]
(b) Tension,

\[T = \frac{Ma}{2} = \frac{M}{2} \times \frac{2g}{3} = \frac{Mg}{3}\]
(c)

Let T' = resultant of tensions
\[\therefore T' = \sqrt{T^2 + T^2} = \sqrt{2}T\]
\[ \therefore T' = \sqrt{2}T = \frac{\sqrt{2}Mg}{3}\]
\[\text{Again, }\tan\theta = \frac{T}{T} = 1\]
\[ \Rightarrow \theta = 45^\circ\]
So, it is `(sqrt2"Mg")/3` at an angle of 45° with the horizontal.
That is the force exerted by the clamp.
APPEARS IN
संबंधित प्रश्न
A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block by another string. Find the tensions in the two strings. Take g = 10 m/s2.
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
A force \[\vec{F} = \vec{v} \times \vec{A}\] is exerted on a particle in addition to the force of gravity, where \[\vec{v}\] is the velocity of the particle and \[\vec{A}\] is a constant vector in the horizontal direction. With what minimum speed, a particle of mass m be projected so that it continues to move without being defelected and with a constant velocity?
Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in the following figure. Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?

Two bodies A and B of same mass are moving with velocities v and 2v, respectively. Compare their (i) inertia and (ii) momentum.
How can Newton's first law of motion be obtained from the second law of motion?
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate: The velocity acquired by the body.
State the magnitude and direction of the force of gravity acting on the body of mass 5 kg. Take g = 9.8 m s-2.
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.
A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The depth of water surface
Multiple Choice Question. Select the correct option.
A force acts on a body of mass 3 kg such that its velocity changes from 4 ms−1 to 10 ms−1. The change in momentum of the body is
Why is it advantageous to turn before taking a long jump?
Name the physical quantity which equals the rate of change of linear momentum.
A stone is thrown vertically upward with a velocity of 9.8 m/s. When will it reach the ground?
A ball is thrown upward and reaches a maximum height of 19.6 m. Find its initial speed?
The INCORRECT statement about Newton's second law of motion is
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is ______.
Why is catching a slow-moving ball easier than catching a fast-moving ball?
