Pulling force making an angle θ to the horizontal is applied on ...
Explanation:
- Free Body Diagram (FBD):
- Draw the free body diagram of the block.
- Show all forces acting on the block including weight, normal reaction, frictional force, and the applied force at an angle θ to the horizontal.
- Resolving Forces:
- Resolve the forces acting on the block into horizontal and vertical components.
- The horizontal component of the applied force will be Fcosθ, and the vertical component will be Fsinθ.
- Force of Friction:
- The maximum force of static friction is given by fs = μsN, where μs is the coefficient of static friction and N is the normal reaction.
- The normal reaction N is equal to the weight W acting downwards.
- Therefore, the maximum force of static friction is fs = μsW.
- The force of friction is given by f = μkN, where μk is the coefficient of kinetic friction.
- Equilibrium Condition:
- For the block to move, the horizontal component of the applied force must overcome the force of friction.
- The force of friction is given by f = μkN = μkW.
- Therefore, Fcosθ = μkW.
- Solving for F, we get F = μkW/cosθ.
- Angle of Friction:
- The angle of friction α is related to the coefficients of friction by tanα = μs = μk.
- Final Expression:
- Substituting the value of μk from the angle of friction relation, we get F = Wsinα/cos(θ-α).
Therefore, the magnitude of force required to move the body is equal to Wsinα/cos(θ-α), which corresponds to option 'C'.