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A sphere of weight w=100N is kept stationary on a rough inclined plane(angle of inclination 30 degrees with the ground) by a horizontal string AB as shown in figure.Find :

1)tension in the string.

2)force of friction on the sphere and

3)normal reaction on the sphere by the plane.

1031_81030_askiitians-2.JPG

Sameer Gupta , 12 Years ago
Grade 12th Pass
anser 1 Answers
Askiitians Tutor Team

To solve this problem, we need to analyze the forces acting on the sphere that is resting on the inclined plane. The sphere has a weight of 100 N, and the incline makes an angle of 30 degrees with the horizontal. The horizontal string AB is keeping the sphere stationary, which means we can apply the principles of equilibrium to find the tension in the string, the force of friction, and the normal reaction force. Let's break this down step by step.

1. Finding the Tension in the String

First, we need to resolve the weight of the sphere into two components: one parallel to the inclined plane and the other perpendicular to it. The weight (W) can be resolved as follows:

  • Weight component parallel to the incline: W_parallel = W * sin(θ)
  • Weight component perpendicular to the incline: W_perpendicular = W * cos(θ)

Substituting the values:

  • W_parallel = 100 N * sin(30°) = 100 N * 0.5 = 50 N
  • W_perpendicular = 100 N * cos(30°) = 100 N * (√3/2) ≈ 86.6 N

Since the sphere is stationary, the tension in the string (T) must balance the component of the weight acting down the incline. Therefore, we have:

T = W_parallel = 50 N

2. Calculating the Force of Friction

The force of friction (F_friction) acts up the incline to prevent the sphere from sliding down. In equilibrium, the sum of forces along the incline must equal zero. Thus, we can express this as:

F_friction + T = W_parallel

Substituting the known values:

F_friction + 50 N = 50 N

From this, we can solve for the force of friction:

F_friction = 50 N - 50 N = 0 N

This indicates that there is no frictional force acting on the sphere because the tension in the string is sufficient to balance the component of the weight acting down the incline.

3. Determining the Normal Reaction Force

The normal reaction force (N) acts perpendicular to the inclined plane. It balances the perpendicular component of the weight. Therefore, we can express this as:

N = W_perpendicular

Substituting the value we calculated earlier:

N = 86.6 N

Summary of Results

To summarize the findings:

  • Tension in the string (T): 50 N
  • Force of friction (F_friction): 0 N
  • Normal reaction force (N): 86.6 N

This analysis shows how the forces interact on an inclined plane and how tension in a string can effectively balance the forces acting on an object. If you have any further questions or need clarification on any part of this, feel free to ask!

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