# Conveyor Belt Friction Problem

- Jan 20, 2019-

1. When the object and the conveyor belt are relatively stationary, there may be static friction or no friction. The specific situation is determined by the motion of the object and the force situation.

Example 1. As shown in the figure, an object of mass m is placed on a horizontal conveyor belt as the conveyor belt moves to the right, and the frictional force of the object is sought under the following circumstances. 1 With a horizontal conveyor belt moving at a constant speed; 2 with the horizontal conveyor belt horizontally to the right with acceleration magnitude a uniform acceleration motion; 3 with the horizontal conveyor belt horizontal to the right with acceleration magnitude a even deceleration motion.

Analysis: 1 When the object moves linearly to the right with the conveyor belt horizontally, as shown in the figure, it is only subjected to gravity and dynamic support force, and is in the balance of two forces; at this time, it is not affected by friction.

2 At this time, the vertical direction is only subject to gravity and support force, and is in the balance of two forces; but it is known from the law of mechanical Newton's motion that there is a static friction force to the right in the horizontal direction, the size of which is f=ma, and the force is as shown in the figure. 3 At this time, the vertical direction is only subject to gravity and support force, and is in the balance of two forces; but there is a static friction force to the left in the horizontal direction, the size of which is f=ma, as shown in the figure.

Example 2. As shown, an object of mass m is placed at an oblique angle on the conveyor belt to move along with the conveyor belt. Try to get the friction of the object under the following scenarios. 1 With the conveyor belt moving at a constant speed, 2 with the conveyor belt upwards to accelerate the movement with the acceleration a, and 3 with the conveyor belt to accelerate the movement with the acceleration a.

Analysis:

1 When the conveyor moves at a constant speed, the force is as shown in the figure.

It can be seen from the equilibrium condition of the object that the static friction force is upward along the slope, and the size f=mgsinα. Independent of the direction of motion of the block

2 As the conveyor belt accelerates upwards with acceleration a, the force is as shown in the figure, which is obtained by the law of mechanical Newton's motion, f-mgsinα=ma, and the static friction force f=mgsinα+ma.

If the object is decelerating down with the acceleration a along with the conveyor belt, the same situation is the same.

3 As the horizontal conveyor belt accelerates downwards with the acceleration level a, at this time, since the relationship between a and gsinα is uncertain, the magnitude and direction of the static friction force f cannot be determined, so the force is as shown in the figure (the direction of f is undetermined) .

If a=gsinα, then f=0

For example, a>gsinα, it is known from the law of mechanical Newton's motion that the direction of the static friction force f is downward along the oblique direction, and there are mgsinα+f=ma, f=ma-mgsinα.

For example, a<gsinα, it is known from the law of mechanical Newton's motion that the direction of the static friction force f is upward along the oblique direction, and there are mgsinα-f=ma, f=mgsinα-ma. 