Question

A 10 kg steel ball is suspended by two string as shown. The tension in the two string are: (g =9.8m/s2)

Diagram is in attachment:- please help me out

Answer 1

see free body diagram,

Let T is tension in left string and T' is tension in right string as shown in figure.

at equilibrium,

upward force = downward force

Tsin45° + T'sin60° = mg

or, T/√2 + √3T'/2 = mg

or, √2T + √3T' = 2mg

where m = 10kg

so, √2T + √3T' = 2 × 10 × 10 = 200N ....(1)

forward force = backward force

Tcos45° = T'cos60°

or, T/√2 = T'/2

or, √2T = T' ......(2)

putting equation (2) in equation (1),

√2T + √3 × √2T = 200

or, (√2 + √6)T = 200

or, T = 200/(√2 + √6) = 200(√6 - √2)/(6 - 2)

= 50(√6 - √2)N

and T' = √2T = 50(2√3 - 2) = 100(√3 - 1)N

hence, answer is 50(√6 - √2)N and 100(√3 - 1)N.