FRICTION
EQUILIBRIUM OF BODY ON INCLINED PLANE
CASE 1: Force Acting Along The Inclined Plane:
1(a) : When the Body is just on the point of motion down the plane:
Consider a block of weight W is kept on an inclined plane having angle of inclination α with horizontal. The block is just at the point of sliding downward. The force acting on the block, which keep it in equilibrium are:
- Weight of the block, W acting vertically downward.
- Applied force, P
- Frictional force, F acting upward along the plane
- Normal reaction, R
Resolving the forces perpendicular to the inclined plane
R = W cos α
Resolving the forces parallel to the plane
P + F = W sin α
P + μR = W sin α
P + μ.W cos α = W sin α
P = W sin α – μ.W cos α
P = W ( sin α – μ cos α)
P = W ( sin α – tan Φ.cos α)
P = W { sin α – [frac up=”sin Φ” down=”cos Φ”] cos α}
P = W [frac up=”( sin α. cos Φ – sin Φ. cos α)” down=”cos Φ”]
P = W. [frac up=”sin (α – Φ)” down=”cos Φ”]
1(b) : When the Body is just on the point of motion up the plane:
Consider a block of weight W is kept on an inclined plane having angle of inclination α with horizontal. The block is just at the point of sliding upward. The force acting on the block, which keep it in equilibrium are:
- Weight of the block, W acting vertically downward.
- Applied force, P
- Frictional force, F acting downward along the plane
- Normal reaction, R
Resolving the forces perpendicular to the inclined plane
R = W cos α
Resolving the forces parallel to the plane
P = W sin α + F
P = W sin α + μR
P= W sin α + μ.W cos α
P = W sin α + μ.W cos α
P = W ( sin α + μ cos α)
P = W ( sin α – tan Φ.cos α)
P = W { sin α + (sin Φ/ cos Φ) cos α}
P = W( sin α. cos Φ + sin Φ. cos α)/ cos Φ
P = W. sin (α + Φ)/ cos Φ
CASE 2: Force Acting Horizontally:
2(a) : When the Body is just on the point of motion down the plane:
Consider a block of weight W is kept on an inclined plane having angle of inclination α with horizontal. The block is just at the point of sliding downward. The force acting on the block, which keep it in equilibrium are:
- Weight of the block, W acting vertically downward.
- Applied force, P acting horizontally
- Frictional force, F acting upward along the plane
- Normal reaction, R
Resolving the forces perpendicular to the inclined plane
R = W cos α + P sin α
Resolving the forces parallel to the plane
P cos α + F = W sin α
P cos α + μR = W sin α
μR = W sin α – P cos α
Putting the value of R, we get
μ(W cos α + P sin α) = W sin α – P cos α
μW cos α + μP sin α = W sin α – P cos α
P cos α + μP sin α = W sin α – μW cos α
P (cos α + μ sin α) = W (sin α – μcos α)
P (cos α + tan Φ.sin α) = W (sin α – tan Φ.cos α)
P {cos α + [frac up=”sin Φ” down=”cos Φ”] sin α} = W {sin α – [frac up=”sin Φ” down=”cos Φ”] cos α}
P(cos α.cos Φ + sin α.sin Φ} = W {sin α.cos Φ – cos α.sin Φ}
P cos (α – Φ)=W sin(α – Φ)
P = W. [frac up=”sin(α – Φ)” down=”cos (α – Φ)”]
P = W tan (α – Φ)
2(b) : When the Body is just on the point of motion up the plane:
Consider a block of weight W is kept on an inclined plane having angle of inclination α with horizontal. The block is just at the point of sliding upward. The force acting on the block, which keep it in equilibrium are:
- Weight of the block, W acting vertically downward.
- Applied force, P acting horizontally
- Frictional force, F acting downward along the plane
- Normal reaction, R
Resolving the forces perpendicular to the inclined plane
R = W cos α + P sin α
Resolving the forces parallel to the plane
P cos α = F + W sin α
P cos α = μR + W sin α
μR = P cos α – W sin α
Putting the value of R, we get
μ(W cos α + P sin α) = P cos α – W sin α
μW cos α + μP sin α = P cos α – W sin α
P cos α – μP sin α = W sin α + μW cos α
P (cos α – μ sin α) = W (sin α + μcos α)
P (cos α – tan Φ.sin α) = W (sin α + tan Φ.cos α)
P {cos α – [frac up=”sin Φ” down=”cos Φ”] sin α} = W {sin α + [frac up=”sin Φ” down=”cos Φ”] cos α}
P(cos α.cos Φ – sin α.sin Φ} = W {sin α.cos Φ + cos α.sin Φ}
P cos (α + Φ)=W sin(α + Φ)
P = W. [frac up=”sin(α + Φ)” down=”cos (α + Φ)”]
P = W tan (α + Φ)
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