A 0.18 kg ball is placed in a shallow wedge with an opening angle of 110° For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume the system is frictionless.
I will assume that the axis of the wedge is vertically upward. The force on the ball at each contact point is directed toward the center of the sphere at each contact point, because there is no friction. The contact force angle is inclined (180-110)/2 = 35 degrees to the horizontal. A vertical force balance tells you that the two contact forces F are equal in magnitude (due to symmetry) and that
2F sin35 = 2F cos55 = M g
To find the force exerted on the ball at each contact point with the wedge, we can use the principles of forces in equilibrium. Since the system is frictionless, the only forces acting on the ball are its weight and the normal force.
1. Determine the weight of the ball:
The weight of an object can be calculated using the formula:
weight = mass * acceleration due to gravity
In this case, the mass of the ball is 0.18 kg and the acceleration due to gravity is approximately 9.8 m/s^2.
So, the weight of the ball is:
weight = 0.18 kg * 9.8 m/s^2 = 1.764 N
2. Calculate the normal force at each contact point:
The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the ball.
So, at each contact point, the normal force exerted on the ball is also 1.764 N.
Therefore, the force exerted on the ball at each contact point with the wedge is 1.764 N.
To determine the force exerted on the ball at each contact point on the wedge, we need to consider the gravitational force acting on the ball and the normal force exerted by the wedge.
1. Gravitational force (weight):
The weight of the ball can be calculated using the formula: weight = mass x gravitational acceleration.
Given that the mass of the ball is 0.18 kg and the gravitational acceleration is approximately 9.8 m/s², we can calculate the weight of the ball:
Weight of the ball = 0.18 kg x 9.8 m/s² = 1.764 N.
2. Normal force:
The normal force is the force exerted by a surface perpendicular to the surface of contact. In this case, the wedge exerts a normal force on the ball at each contact point.
Since the system is frictionless, the normal force is equal to the weight of the ball. Therefore, the normal force exerted on the ball at each contact point is 1.764 N.
So, the force exerted on the ball at each contact point is 1.764 N.