A block of weight 400N resting on a rough horizontal floor supports a block B weighing 200N. The blocks are connected to a string passing over a smooth pulley as shown in figure. Determine the least horizontal force P to be applied to the block A so as to just move it towards right. Coefficient of friction is 0.25
Answers
Given:
A block of weight 400N resting on a rough horizontal floor supports a block B weighing 200N. The blocks are connected to a string passing over a smooth pulley.
Coefficient of friction is 0.25.
To find:
The least horizontal force P to be applied to the block A so as to just move it towards right.
Solution:
Let m1 = 400 N and m2 = 200 N
Tension T and the opposite force P.
Using Newton's second law, we have,
∑ Fx = m1 × ax
T - 0.25N1 = 0
∑ Fy = m1 × ay
N1 - 200N = 0
⇒ N1 = 200N
T - 0.25N1 = 0
T = 0.25N1
T = 0.25 × 200
⇒ T = 50N
∑ Fx = m2 × ax
T - P + 0.25N1 + 0.25N2 = 0
∑ Fy = m2 × ay
N2 - 400N - N1 = 0
N2 = 400N + N1
N2 = 400N + 200N
⇒ N2 = 600N
T - P + 0.25N1 + 0.25N2 = 0
50 - P + 0.25 × 200 + 0.25 × 600 = 0
50 - P + 50 + 150 = 0
⇒ P = 250N
The least horizontal force is P = 250 N
Step-by-step explanation:
Let m1 = 400 N and m2 = 200 N
Tension T and the opposite force P.
Using Newton's second law, we have,
∑ Fx = m1 × ax
T - 0.25 N1 = 0
∑ Fy = m1 × ay
N1 - 200 N = 0
⇒ N1 = 200 N
T - 0.25N1 = 0
T = 0.25 N1
T = 0.25 × 200
⇒ T = 50 N
∑ Fx = m2 × ax
T - P + 0.25 N1 + 0.25 N2 = 0
∑ Fy = m2 × ay
N2 - 400 N - N1 = 0
N2 = 400 N + N1
N2 = 400 N + 200 N
⇒ N2 = 600 N
T - P + 0.25 N1 + 0.25 N2 = 0
50 - P + 0.25 × 200 + 0.25 × 600 = 0
50 - P + 50 + 150 = 0
⇒ P = 250 N
The least horizontal force is P = 250 N