If friction is present between the shaft and pulley, the system will have a lesser acceleration than when it has no friction. This is because friction is a force that opposes motion and can reduce the overall net force acting on the system.
When there is no friction, the net force acting on the system is only the force applied to the pulley, which is the tension in the string. However, when friction is present, there is an additional force acting on the system that opposes its motion, leading to a decrease in the overall net force.
This decrease in net force results in a decrease in acceleration, as acceleration is directly proportional to the net force acting on the system. Therefore, the presence of friction between the shaft and pulley will lead to a slower acceleration of the system compared to when there is no friction.
A ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass. The block lies on a frictionless incline of angle.
a. Find the magnitude of the acceleration of the two objects and the tension in the
cord.
b. What happens in this situation if the angle = 90O?
c. What happens if the mass m1 = 0?
a. To find the magnitude of the acceleration of the two objects and the tension in the cord, we need to draw a free body diagram for each object.
For the block:
There is a gravitational force (mg) acting downwards.
There is a normal force (N) acting perpendicular to the incline.
There is a tension force (T) acting upwards and parallel to the incline.
There is no friction force since the incline is frictionless.
We can resolve the gravitational force into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ). The net force acting on the block along the incline is given by:
Fnet = m2 * a = T - m2 * g * sinθ
For the ball:
There is a gravitational force (m1 * g) acting downwards.
There is a tension force (T) acting upwards.
The net force acting on the ball is given by:
Fnet = m1 * a = T - m1 * g
Since the cord is lightweight, we can assume that the tension is the same for both objects. Therefore, we can set T = T in the two equations above, and solve for a and T:
m2 * a = T - m2 * g * sinθ
m1 * a = T - m1 * g
Adding the two equations gives:
(m1 + m2) * a = T - (m1 * g + m2 * g * sinθ)
Solving for a:
a = (T - (m1 + m2) * g * sinθ) / (m1 + m2)
To find T, we can substitute the above expression for a into one of the equations:
T = m2 * g * sinθ + m1 * g
Therefore, the magnitude of the acceleration of the two objects is given by:
a = (m2 * g * sinθ - m1 * g) / (m1 + m2)
And the tension in the cord is given by:
T = m2 * g * sinθ + m1 * g
b. If the angle θ = 90°, then the incline becomes a vertical wall. In this case, the block will not move since there is no net force acting on it along the wall. The ball will fall freely under the influence of gravity.
c. If m1 = 0, then the ball has no mass and will fall freely under the influence of gravity. The block will still experience a net force along the incline due to its weight, and will accelerate downwards with an acceleration given by:
a = m2 * g * sinθ / m2 = g * sinθ
The tension in the cord will be zero since there is no force acting on the cord due to the massless ball.