a uniform rod with a mass 120g and a length of 130cm is suspended by a wire from a point 80cm from the rod's left and. what mass must be hang from the right end of the rod it to be in equlibrium ?

Question

what will be the tension of the wire ?

Answer 1

To find the mass that must be hung from the right end of the rod, we can use the principle of moments.

First, let's calculate the torque exerted by the rod about the point of suspension:

Torque = Force x Distance

Since the rod is in equilibrium, the torque on the left side of the point of suspension must balance the torque on the right side.

Torque on the left side (due to rod) = 120g x 80cm = 9600 g cm
Torque on the right side (due to hanging mass) = Mass x 50cm (distance of hanging mass from point of suspension)

Setting the two torques equal:

9600 = Mass x 50

Mass = 9600 / 50
Mass = 192g

Therefore, a mass of 192g must be hung from the right end of the rod for it to be in equilibrium.

To find the tension in the wire, we can consider the vertical forces acting on the rod:

Tension in the wire = Weight of rod + Weight of hanging mass

Weight of rod = 120g x 9.81 m/s^2 = 1.176 N
Weight of hanging mass = 192g x 9.81 m/s^2 = 1.882 N

Therefore, Tension in the wire = 1.176 N + 1.882 N
Tension in the wire = 3.058 N

So, the tension in the wire will be 3.058 N.