A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, 1 and 2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of 2 is 6 times that of 1. Find the distances a and b that locate the axis. Note that a and b are not drawn to scale.
a=
b=
To find the distances a and b that locate the axis, we need to use the concept of torque.
Torque is the rotational equivalent of force, and it is calculated as the product of the force and the distance from the point of rotation (axis) to the line of action of the force. Mathematically, torque (τ) is given by the equation τ = r x F, where r is the distance vector from the axis to the line of action of the force, and F is the force vector.
In this case, we are given that the net torque produced by the forces is zero. This means that the sum of the torques produced by both forces must be zero.
Let's analyze the torques produced by each force in both parts a and b:
In part a:
- Force 1: The torque produced by force 1 is zero because the line of action of the force passes through the axis. Therefore, the distance a does not affect the torque and can be any value.
- Force 2: The torque produced by force 2 is also zero because the line of action of the force passes through the axis. Therefore, the distance b does not affect the torque and can be any value.
In summary, the distances a and b can be any values, as long as they ensure that the line of action of force 1 and force 2 passes through the axis.
However, in part b, we are given a specific condition: the magnitude of force 2 is 6 times that of force 1.
Let's use this condition to find the relationship between a and b:
- Force 1: Let's assume the magnitude of force 1 is F. It has a line of action passing through the axis, so its torque is zero.
- Force 2: The magnitude of force 2 is 6F. It also has a line of action passing through the axis, so its torque is zero.
Since the torque produced by force 2 is given by τ = r x F, and the torque is zero, we can write:
r x 6F = 0
Since the cross product of two vectors is zero when the vectors are parallel or one of them is the zero vector, we can deduce that r must be parallel to force 2.
Therefore, a and b can be any values, as long as the line of action of force 1 passes through the axis, and the line of action of force 2 is parallel to it (since force 2 has a magnitude 6 times that of force 1).
In conclusion:
a can be any value, as long as the line of action of force 1 passes through the axis.
b can be any value, as long as it ensures that the line of action of force 2 is parallel to the line of action of force 1.