A metal rod of length 90cm has a disc of radius 24cm fixed rigidly at its centre,and the assembly is pivoted from its centre.
Two forces each of magnitude 30N,are applied normal to the rod at each end so as to produce a turning effect.A rope is attached to the edge of the disc to prevent rotation.
calc.
a)the torque of the couple produced by the 30N forces.
b)The tension T in the rope.
To calculate the torque of the couple produced by the 30N forces, we can use the formula for torque: torque = force x perpendicular distance from the pivot. In this case, the perpendicular distance is half the length of the rod, since the disc is fixed at the center.
a) To find the torque, we can calculate it for one of the forces and then multiply it by 2 since there are two forces.
The perpendicular distance from the pivot to one end of the rod is 90cm/2 = 45cm = 0.45m.
So, the torque produced by one force is torque = force x perpendicular distance = 30N x 0.45m = 13.5Nm.
Since there are two forces, the total torque would be 2 x 13.5Nm = 27Nm.
b) To find the tension T in the rope, we need to consider the equilibrium of forces. The forces applied by the 30N forces create a clockwise torque, while the tension in the rope creates an equal and opposite counterclockwise torque to prevent rotation.
The torque created by the tension in the rope is given by torque = force x perpendicular distance from the pivot. Since the force applied by the tension is perpendicular to the radius of the disk, the perpendicular distance is equal to the radius of the disk.
The torque created by the tension is equal to the torque created by the 30N forces, so we can write:
T x 24cm = 27Nm
First, we need to convert the radius to meters: 24cm = 0.24m.
So the equation becomes:
T x 0.24m = 27Nm
Now we can solve for T:
T = 27Nm / 0.24m = 112.5N
Therefore, the tension T in the rope is 112.5N.