Question

A uniform horizontal metre scale of mass M is suspended by two vertical strings attached to its two ends.A block of mass 3M is placed on the 60 centimetre mark the tension in the two strings are in the ratio​

Answer 1

weight of horizontal metre scale = Mg(acts on middle point of scale )

weight of block = 3Mg(act 60cm from end A )

now, torque at middle point of metre scale = 0 [ because, there doesn't have any rotational motion ]

so, torque due to end A + torque due to metre scale + torque due to block + torque due to end B = 0

or, T × 50cm(clockwise) + Mg × 0 + 3Mg × 10(clockwise) + T' × 50cm(anticlockwise) = 0

or, T × 50 + 3Mg - T' × 50= 0

or, 5T' - 5T= 3Mg .....(1)

also at equilibrium,

T + T' = Mg + 3Mg = 4Mg

or, 5T + 5T' = 20Mg .......(2)

from equation (1) and (2),

10T' = 23Mg

or, T' = 2.3Mg and T = 4Mg - 2.3Mg = 1.7Mg

so, T'/T = 2.3Mg/1.7Mg = 23/17