To find the mass that must be hung from the right end of the rod in order for it to be in equilibrium, we can set up the following equation:
Sum of torques = 0
The torque due to the hanging mass on the right end of the rod will be:
Torque = (mass) x (g) x (distance from pivot point)
= M x 9.8 x 50cm
= M x 9.8 x 0.5m
The torque due to the uniform rod itself will be:
Torque = (mass of rod) x (g) x (center of mass distance from pivot point)
= 0.12kg x 9.8 x 0.65m
Since the rod is in equilibrium, the two torques must balance each other out:
M x 9.8 x 0.5 = 0.12 x 9.8 x 0.65
Solving for M, we get:
M = (0.12 x 0.65) / 0.5
M = 0.156 kg
Therefore, a mass of 0.156 kg must be hung from the right end of the rod for it to be in equilibrium.
To find the tension of the wire, we can use the equation for the sum of forces in the vertical direction at equilibrium:
Tension in wire = weight of rod + weight of hanging mass
T = (mass of rod x g) + (mass hung from right end x g)
T = (0.12 x 9.8) + (0.156 x 9.8)
T = 1.176 + 1.5288
T = 2.7048 N
The tension in the wire will be 2.70 N.
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 ?
what will be the tension of the wire ?
1 answer