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A body of mass 10kg is placed on an inclined surface of angle 30 . If the coefficient of limiting friction is 1/(3)½ find the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane. Please give answer with free body diagram also.?
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A body of mass 10kg is placed on an inclined surface of angle 30 . If ...
Free Body Diagram

To solve this problem, let's start by drawing a free body diagram of the body on the inclined plane.

1. Draw a diagram of the inclined plane with the body placed on it.
2. Draw a normal force vector perpendicular to the inclined plane, acting upwards.
3. Draw the weight vector acting vertically downwards, with magnitude mg, where m is the mass of the body and g is the acceleration due to gravity.
4. Draw the friction force vector parallel to the inclined plane, opposing the direction of motion.
5. Draw the force vector parallel to the inclined plane, representing the force required to push the body up the inclined plane.

Analysis of Forces

Now, let's analyze the forces acting on the body:

1. Normal Force (N): The normal force is the force exerted by the inclined plane on the body perpendicular to the plane's surface. It is equal in magnitude and opposite in direction to the component of the weight that is perpendicular to the inclined plane. In this case, N = mgcosθ, where θ is the angle of inclination.

2. Weight (W): The weight of the body is the force exerted on the body due to gravity. It is equal to mg, where m is the mass of the body and g is the acceleration due to gravity.

3. Friction Force (f): The friction force is the force that opposes the motion of the body on the inclined plane. It is given by the equation f = μN, where μ is the coefficient of friction and N is the normal force. In this case, μ = 1/(3)½ and N = mgcosθ. Therefore, f = (1/(3)½)(mgcosθ).

4. Force Required to Push the Body (F): The force required to push the body up the inclined plane is the force needed to overcome the friction force. Since the force is parallel to the inclined plane, it is equal in magnitude to the friction force. Therefore, F = (1/(3)½)(mgcosθ).

Calculation

In this problem, the mass of the body is given as 10 kg and the angle of inclination is 30°. Therefore, we can substitute these values into the equation for the force required to push the body:

F = (1/(3)½)(mgcosθ)
= (1/(3)½)(10 kg)(9.8 m/s²)(cos30°)
= (1/(3)½)(10)(9.8)(√3/2)
= (1/(3)½)(10)(9.8)(√3/2)
≈ 85.17 N

Therefore, the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane, is approximately 85.17 N.
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A body of mass 10kg is placed on an inclined surface of angle 30 . If the coefficient of limiting friction is 1/(3)½ find the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane. Please give answer with free body diagram also.?
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A body of mass 10kg is placed on an inclined surface of angle 30 . If the coefficient of limiting friction is 1/(3)½ find the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane. Please give answer with free body diagram also.? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body of mass 10kg is placed on an inclined surface of angle 30 . If the coefficient of limiting friction is 1/(3)½ find the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane. Please give answer with free body diagram also.? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 10kg is placed on an inclined surface of angle 30 . If the coefficient of limiting friction is 1/(3)½ find the force required to just push the body up the inclined plane, if the force is parallel to the inclined plane. Please give answer with free body diagram also.?.
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