Two bodies of weights 50N and 10N are connected to the two ends of a light inextensible string which passes over a smooth pulley. The weight 10N is placed on a rough inclined plane of angle of inclination 20° while the weight of 50N hangs vertically downward. If the coefficient of friction between the body and he plane is 0.2, make calculations for the acceleration of the system, tension induced in the string, action at the pulley and the distance moved by the body in 3 seconds starting from rest.
To solve this problem, we can divide it into two parts: the motion of the 10N body on the inclined plane and the motion of the 50N body hanging vertically.
1. Motion of the 10N body on the inclined plane:
The weight of the 10N body can be resolved into two components: one parallel to the inclined plane (10N * sin 20°) and one perpendicular to the inclined plane (10N * cos 20°). The force of friction can be calculated as the product of the normal force (10N * cos 20°) and the coefficient of friction (0.2).
The net force parallel to the inclined plane is given by:
Fnet = m * a
10N * sin 20° - friction = 10N * a
Simplifying this equation, we get:
10N * sin 20° - 0.2 * (10N * cos 20°) = 10kg * a
Solving for a, we find:
a = (10N * sin 20° - 0.2 * (10N * cos 20°)) / 10kg
2. Motion of the 50N body hanging vertically:
The weight of the 50N body is equal to its mass times the acceleration due to gravity (50N = 50kg * 9.8m/s^2). Since the body is hanging vertically, the tension in the string is equal to the weight of the body (50N).
Now, we can calculate the acceleration of the system by considering the net force and mass of the system:
50N - tension = (10kg + 50kg) * a
Simplifying this equation, we find:
50N - 50N = 60kg * a
Since the net force on the system is zero, the acceleration of the system is zero.
Now, let's calculate the distance moved by the body in 3 seconds starting from rest using the equation:
S = ut + (1/2)at^2
For the motion of the 10N body on the inclined plane:
u = 0 (starting from rest)
a = (10N * sin 20° - 0.2 * (10N * cos 20°)) / 10kg (calculated earlier)
t = 3s
Plugging in these values, we find:
S1 = 0 + 0.5 * (10N * sin 20° - 0.2 * (10N * cos 20°)) / 10kg * (3s)^2
For the motion of the 50N body hanging vertically:
u = 0 (starting from rest)
a = 0 (calculated earlier)
t = 3s
Plugging in these values, we find:
S2 = 0 + 0.5 * 0 * (3s)^2
Finally, the total distance moved by the body in 3 seconds is given by:
S = S1 + S2
Note: The action at the pulley is the tension induced in the string, which is equal to the weight of the 50N body (50N).