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A body weight of 300 newton is lying on a rough horizontal plane having a coefficient of friction as a 0.3 while acting at an angle of 25 degree with the horizontal plane what is the horizontal force act on the body?
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A body weight of 300 newton is lying on a rough horizontal plane havin...
Sum of all horizontal forces is zero
wsin(25degree) - mue×wcos(25 degree)
answer is 45.217Newton on solving
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A body weight of 300 newton is lying on a rough horizontal plane havin...
Given information:
- Body weight = 300 Newton
- Coefficient of friction (μ) = 0.3
- Angle of friction (θ) = 25 degrees

Calculating the horizontal force:
To determine the horizontal force acting on the body, we need to consider the force of friction. The force of friction is given by the equation:

Friction force (F) = μ * Normal force (N)

Where:
- μ is the coefficient of friction
- N is the normal force

Calculating the normal force:
The normal force is the force exerted by the horizontal plane on the body perpendicular to the plane. In this case, the body is lying on the plane, so the normal force is equal to the weight of the body.

Normal force (N) = Body weight
N = 300 Newton

Calculating the friction force:
Substituting the given values into the equation, we can calculate the friction force:

Friction force (F) = μ * N
F = 0.3 * 300
F = 90 Newton

Resolving the force:
The friction force acts opposite to the direction of motion or impending motion. In this case, the body is lying on the plane and not moving, so the friction force acts horizontally.

Horizontal force = Friction force
Horizontal force = 90 Newton

Therefore, the horizontal force acting on the body is 90 Newton.
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A body weight of 300 newton is lying on a rough horizontal plane having a coefficient of friction as a 0.3 while acting at an angle of 25 degree with the horizontal plane what is the horizontal force act on the body?
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