Answer:
Part 1)
[tex]\tau_1 = 5 \times (0.50) = 2.5 N m[/tex]
Part 2)
[tex]\tau_2 = 14 \times (0.30) = 4.2 N m[/tex]
Part 3)
[tex]\tau_3 = 1.4 N m[/tex]
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
[tex]\tau_1 = 5 \times (0.50) = 2.5 N m[/tex]
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
[tex]\tau_2 = 14 \times (0.30) = 4.2 N m[/tex]
Now torque due to weight of the scale is given as
[tex]\tau_3 = 7(0.20)[/tex]
[tex]\tau_3 = 1.4 N m[/tex]
now torque on left side of scale is given as
[tex]\tau_{left} = \tau_1 + \tau_3[/tex]
[tex]\tau_{left} = 2.5 + 1.4 = 3.9 N m[/tex]
Torque on right Side is given as
[tex]\tau_{right} = \tau_2 = 4.2 Nm[/tex]
Since torque on right side is more so here it will turn and slip over it