Question

A student places blocks on a 100cm long see-saw as shown/

Answer 1

Answer:

Part 1)

$\tau_1 = 5 \times (0.50) = 2.5 N m$

Part 2)

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Part 3)

$\tau_3 = 1.4 N m$

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

$\tau_1 = 5 \times (0.50) = 2.5 N m$

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Now torque due to weight of the scale is given as

$\tau_3 = 7(0.20)$

$\tau_3 = 1.4 N m$

now torque on left side of scale is given as

$\tau_{left} = \tau_1 + \tau_3$

$\tau_{left} = 2.5 + 1.4 = 3.9 N m$

Torque on right Side is given as

$\tau_{right} = \tau_2 = 4.2 Nm$

Since torque on right side is more so here it will turn and slip over it