A see-saw consists of a pivot and a heavy (24 kg) uniform plank which is 6.6 m long. A 12kg box is placed at one end of the plank. How far from the other end should the pivot be so that the plank is balanced?
To find the position of the pivot, we need to ensure that the total torque on both sides of the plank is equal. Torque is the product of the weight (mass multiplied by gravitational acceleration) and the distance from the pivot. In this case, we have two torques, one from the box and one from the plank itself.
First, let's calculate the torque from the box. The weight of the box can be calculated by multiplying its mass (12 kg) by the acceleration due to gravity (9.8 m/s^2). So the weight of the box is 12 kg * 9.8 m/s^2 = 117.6 N.
Since the box is placed at one end of the plank, the distance from the pivot to the box is x (which is what we need to find). Therefore, the torque from the box is given by the product of the weight and the distance, which is 117.6 N * x.
Now, let's calculate the torque from the plank itself. The weight of the plank can be calculated by multiplying its mass (24 kg) by the acceleration due to gravity (9.8 m/s^2). So the weight of the plank is 24 kg * 9.8 m/s^2 = 235.2 N.
The distance from the pivot to the center of the plank is half of its length, which is 6.6 m / 2 = 3.3 m. Therefore, the torque from the plank is given by the product of the weight and the distance, which is 235.2 N * 3.3 m.
Since the see-saw is balanced, the sum of torques on both sides must be zero. Therefore,
117.6 N * x = 235.2 N * 3.3 m
Now we can solve this equation to find the value of x.
Dividing both sides of the equation by 117.6 N, we get:
x = (235.2 N * 3.3 m) / 117.6 N
Simplifying the expression gives:
x = 6.6 m
Therefore, the pivot should be placed 6.6 m from the other end of the plank so that it is balanced.