Question

2. A uniform rod of weight 5N and length 1m is pivoted at a point 20cm from one of its ends. A weight is hung from the other end so that the rod balances horizontally. What is the value of the weight?

Answer 1

To find the value of the weight that needs to be hung from the other end of the rod, we can use the principle of equilibrium and calculate the torque exerted by the rod's weight. By setting the torque equal to the torque exerted by the weight hung on the other end, we can solve for the weight. In this case, the weight is 1.25N.

Step-by-step explanation:

To determine the value of the weight that needs to be hung from the other end of the rod in order to balance it horizontally, we can set up an equation based on the principle of equilibrium. The torque exerted by the weight on one side of the rod must be equal to the torque exerted by the rod's weight on the other side. The torque is calculated by multiplying the force applied by the distance from the pivot point.

First, let's calculate the torque exerted by the rod's weight. The weight of the rod is 5N, and it is located 20cm from the pivot point. So, the torque exerted by the rod is 5N x 20cm = 5N x 0.2m = 1N·m.

Since the rod is balanced horizontally, the torque exerted by the weight hung on the other end must also be 1N·m. To find the value of the weight, we can rearrange the torque equation: torque = force x distance. In this case, the distance is the length of the rod minus the distance of the weight from the pivot point. So, the torque equation becomes: 1N·m = weight x (1m - 20cm). Solving for the weight, we get: weight = 1N·m / (1m - 20cm) = 1N·m / 0.8m = 1.25N.

Therefore, the value of the weight that needs to be hung from the other end of the rod is 1.25N.

Answer 2

The weight is 7.5 N

Step-by-step explanation:

Since the rod is uniform, its weight applies downwards at its middle (i.e 50 cm mark). The diagram attached to this answer shows the appropriate sketch.

Applying the principle of moment to the question, let the weight by represented by W.

Total clockwise moment = Total counterclockwise moment

`F_(1)` x
`d_(1)` =
`F_(2)` x
`d_(2)`

5 x 0.3 = W x 0.2

1.5 = 0.2 W

W =
`(1.5)/(0.2)`

= 7.5 N

The value of the weight is 7.5 N.