A man hangs from a horizontal bar with his arms at right angles to each other. If the weight of the man is 69kg the tension in each arm is 1) 60 kg wt 2) 60√2 3) 30

If we draw fbd for the man we find that mg acts down . If you resolve the tension act on hands you get up ward force is equal to 2Tcosx where x=45°.As he is in rest upward force equal to downward force. 2Tcos45°=690N T√2=690N T=690/√2N T =34.5√2×10N T=34.5√2kgwt