Question

Find the acceleration and tension in the string in the following cases. The string and pulley are mass less​

Answer 1

The tension in the block is T = ( √ 2 – √2 / 3 ) mg = 2 √2 mg / 3

Explanation:

N − 100 = 0

⇒ N = 100  

f = μ N = 0.6 × 100 = 60 N

T − f = 10 a

T − 60 = 10 a

a = - 2 / 3 m/s^2

For block of mass 2 m:

N 1 = 2 mg cos (45∘) = √2 mg

f 1 = μ1 N 1 = √2 / 3 mg

Along the plane: 2 mg sin(45∘) − T − F 1 = 2 ma

√2 mg − T − 2 – √3 mg = 2 ma -----(1)

Tension of block 2 m:

2 mg sin(45∘) = T + f1  

√2 mg = T + √2 / 3 mg

T = ( √ 2 – √2 / 3 ) mg = 2 √2 mg / 3