Two bodies of mass 6 kg and 9 kg are connected with a light inextensile string passing over a smooth pulley with one of the bodies on a smooth inclined plane (θ = 30°) and the other hanging freely from the other end. Show that the time taken by 9 kg to pull 6 kg up the entire length is half the time taken by 6 kg to pull 9 kg over the same plane.

Two bodies of mass 6 kg and 9 kg are connected with a light inextensile string passing over a smooth pulley with one of the bodies on a smooth inclined plane (θ = 30°) and the other hanging freely from the other end. Show that the time taken by 9 kg to pull 6 kg up the entire length is half the time taken by 6 kg to pull 9 kg over the same plane.
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1 Answer

Answer :

Given:

Mass of body A (mA) = 6 kg

Mass of body B (mB) = 9 kg

Angle of inclination of the plane (θ) = 30°

Let's assume that the acceleration of body A (aA) and body B (aB) are in opposite directions, with aA pointing upwards along the inclined plane, and aB pointing downwards due to gravity.

According to Newton's second law of motion, the net force acting on a body is equal to the product of its mass and acceleration (F = ma).

For body A:

The force acting on body A (FA) along the inclined plane is given by:

FA = mA * aA

For body B:

The force acting on body B (FB) due to gravity is given by:

FB = mB * g, where g is the acceleration due to gravity (g = 9.8 m/s^2)

As the string is light and inextensible, the tension in the string (T) will be the same on both sides of the pulley.

Now, considering the motion of body A along the inclined plane, we can resolve the gravitational force (mB * g) of body B into two components: one parallel to the inclined plane (mB * g * sinθ), and one perpendicular to the inclined plane (mB * g * cosθ). The component of gravitational force parallel to the inclined plane (mB * g * sinθ) will act in the direction opposite to the motion of body A, and the component of gravitational force perpendicular to the inclined plane (mB * g * cosθ) will not affect the motion of body A along the inclined plane.

Therefore, the net force acting on body A (FA) along the inclined plane will be the difference between the force due to tension in the string (T) and the component of gravitational force parallel to the inclined plane (mB * g * sinθ):

FA = T - mB * g * sinθ

Similarly, the net force acting on body B (FB) will be the sum of the force due to tension in the string (T) and the component of gravitational force parallel to the inclined plane (mB * g * sinθ):

FB = T + mB * g * sinθ

Now, using the equation FA = mA * aA, we can write:

T - mB * g * sinθ = mA * aA (equation 1)

And using the equation FB = mB * aB, we can write:

T + mB * g * sinθ = mB * aB (equation 2)

Adding equation 1 and equation 2, we get:

2T = (mA + mB) * aA + mB * aB

As the acceleration of body A (aA) and body B (aB) are in opposite directions, we have aA = -aB (since aA is upward along the inclined plane and aB is downward due to gravity).

Substituting this into the above equation, we get:

2T = (mA + mB) * aA - mB * aA

2T = (mA + mB - mB) * aA

2T = mA * aA

T = (mA/2) * aA

Now, we know that the time taken (t) is given by the formula:

t = √(2d/a), where d is the distance travelled
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