Question

Refer to figure.If the co-efficient of friction is 0.60 for all contact surfaces and θ = 30o,what force P applied to the block B acting down and parallel to the incline will start motion and what will be the tension in the cord parallel to inclined plane attached to A.
Take WA=120 N and WB=200 N.

Answer 1

Given : : μ=0.6
θ = 30°
W_A = 120 N
W_B = 200 N
To find : Force P
Solution :
F₁ = μN₁ = 0.6N₁ ………(1)
F₂ = μN₂ = 0.6N₂ ……….(2)
Consider FBD of block A

The block is considered to be in equilibrium
Applying conditions of equilibrium
ΣFy = 0
N₁ - 120cos30 = 0
N₁ = 103.923 N ……….(3)
From (1)
F₁ = 0.6 x 103.923
= 62.3538 N
Applying conditions of equilibrium
ΣFx = 0
F₁ + 120sin30 – T = 0
T = 122.3538 N
Consider FBD of block B
Applying conditions of equilibrium
ΣFy = 0
N₂ - N₁ - 200cos30 = 0
F₂ = 0.6 x 277.1281
= 166.2769 N               From (2)
Applying conditions of equilibrium
ΣFx = 0
P - F₁ - F₂ + 200sin30 = 0
P = 128.6307 N
Force required on block B to start the motion is 128.6307 N Tension T in the cord parallel to inclined plane attached to A=122.3538 N