Question

A mass M of 100 kg is suspended with use of strings A. B and C as shown in fig, where W is vertical
wall and Ris a rigid horizontal rod. The tension in string B is
90° T7
/
в
1а
A su
4 M = 100 kg
to
100
(A) 100g N
(B) 0
(C) 100/2 g N
(D)
ON​

Answer 1

Answer:

100g N

Explanation:

Draw the FBD of the block ,

then equate all opposite forces

T/√2 = 100g

T = 100√2g

t = T/√2

=100g

Answer 2

Answer:

100g N.

Explanation:

Since, the mass is hanging from the string at point A so, the weight of the body will be mg=100g. Now, Tsin45 will be 100g which on solving we will get T=100g√2. We get the tension on taming the components of the tension T.

Now, this tension will be equal to T'cos45. Which on solving we will get the value of T' as 100g. The value of the T'cos45 is determined by the taking the component of the tension on the line.