Question

Determine the force P required to move the block A of 5000 N weight up the inclined plane, coefficient of friction between all contact surfaces is 0.25. Neglect the weight of the wedge and the wedge angle is 15 degrees.

Given : Weight of block A = 5000 N
μ_s=0.25
Wedge angle = 15^º

To find : Force P required to move block A up the inclined plane

Answer 1

The impending motion of block A is to move up
The block A is in equilibrium
N_1,N₂,N₃ are the normal reactions
Fs₁ = μ1N₁ = 0.25N₁
F_s2 = μ₂N₂ = 0.25N₂
Fs₃ = μ₃N₃ = 0.25N₃
Applying the conditions of equilibrium
ΣFy = 0
∴ -5000 + N₁ cos 60 – Fs₁ sin 60 – Fs₂ sin 15 + N₂ cos 15 = 0
∴ N₁ x 0.5 – 0.25N₁ x 0.866 -0.25 N₂ x 0.2588 + N₂ x 0.9659 = 5000 (From 1)
∴ 0.2835 N₁ +0.9012 N₂ = 5000 ……………..(2)
Applying the conditions of equilibrium
ΣFx = 0
∴N₁ sin 60 +Fs₁ cos 60 –Fs₂ cos 15 – N₂ sin 15 =0
∴ 0.866 N₁ + 0.25 x N₁ x 0.5 -0.25 x N₂ x 0.9659 –N₂ x 0.2588 = 0(From 1)
∴ 0.991 N₁ -0.5003 N₂ = 0
Solving equation, no 2 and 3
N₁ = 2417.0851 N
N₂ = 4787.79 N
The impending motion of block B is towards left Block B is in equilibrium. Applying the conditions of equilibrium
ΣFy = 0
∴ N₃ + F_s2 sin 15 –N₂ cos 15 = 0
∴N₃ + 0.25N₂ x 0.2588 – N₂ x 0.9659 = 0
∴ N₃ – 0.9012 N₂ = 0
∴ N₃= 0.9012 x 4787.79 = 4314.7563
Applying conditions of equilibrium
ΣFx = 0
∴ -P + F_s3 +F_s2 cos 15 +N₂ sin 5 =0
∴ 0.25 N₃ +0.25 N₂ x 0.9659 +N₂ x 0.2588 =P
∴ P = 0.25 N₃ +0.5003 N₂ = 0.25 X 4314.7563 + 0.5003 x 4787.79 = 3474 N

The force P required to move the block A of weight 5000 N up the inclined plane is P=3474 N