A solid uniform 45.0 kg ball of diameter 32.0 cm is supported against a vertical frictionless wall using a thin 30 cm wire of negligible mass. (a) make a free body diagram for the ball and use it to find the tension in the wire and (b) how hard does the ball push against the wall?
8 hours ago - 4 days left to answer.
Additional Details
It looks as though the 30 cm length coverst the length of the wire from the wall to the surface of the sphere. When I worked it out I added the length of the radius to it.
(a) Free body diagram:
Assuming that the wire is attached to the top of the ball, there are two forces acting on the ball: the weight acting downwards and the tension in the wire acting upwards.
The weight of the ball can be calculated using the formula:
Weight = mass * gravitational acceleration
Weight = 45.0 kg * 9.8 m/s^2 (acceleration due to gravity)
The tension in the wire is equal to the weight of the ball since there is no acceleration in the vertical direction (ball is in equilibrium). So, the tension is also:
Tension = 45.0 kg * 9.8 m/s^2
(b) How hard does the ball push against the wall?
Since the ball is in equilibrium, the net force acting on it must be zero. The ball pushes against the wall with a force equal to the tension in the wire, which is also equal to the weight of the ball.
Therefore, the ball pushes against the wall with a force of 45.0 kg * 9.8 m/s^2.
So, the ball pushes against the wall "as hard as gravity" or "with the force of a 45.0 kg ball under the influence of gravity."
Remember, this answer assumes that the wire is perfectly vertical, the ball is not moving, and there is no other external force acting on it.
(a) To make a free body diagram for the ball, we need to consider all the forces acting on it.
Let's denote the tension in the wire as T and the weight of the ball as mg, where m is the mass of the ball and g is the acceleration due to gravity.
The forces acting on the ball are:
1. Tension in the wire (pointing upwards)
2. Weight of the ball (pointing downwards)
3. Normal force exerted by the wall (pointing towards the wall)
Since the wire is supporting the ball in equilibrium, the tension in the wire balances the weight of the ball. Therefore, T = mg.
(b) To find how hard the ball pushes against the wall, we need to calculate the normal force exerted by the wall on the ball.
The normal force exerted by the wall is equal in magnitude and opposite in direction to the force the ball exerts on the wall. This force is perpendicular to the wall.
In this case, the only force exerted on the wall by the ball is the normal force, and its magnitude is equal to the weight of the ball, which is mg.
Therefore, the ball pushes against the wall with a force of mg.
To find the tension in the wire and how hard the ball pushes against the wall, we can use two key principles: equilibrium and Newton's Third Law.
(a) Free Body Diagram and Tension in the Wire:
In the free body diagram, draw the forces acting on the ball. These include the weight (mg) acting downward, the tension force (T) exerted by the wire, and the normal force (N) exerted by the wall. Since the ball is in equilibrium, the sum of the forces in the vertical direction must equal zero.
Vertical forces:
∑Fy = N - mg = 0
From this equation, we can see that the normal force N is equal to the weight mg.
N = mg
The tension in the wire, T, can be found by considering the horizontal forces:
Horizontal forces:
∑Fx = T = 0
So, the tension in the wire is equal to zero, meaning there is no horizontal force acting on the ball.
(b) How hard the ball pushes against the wall:
According to Newton's Third Law, every action force has an equal and opposite reaction force. In this case, the force exerted by the wall on the ball is equal in magnitude but opposite in direction to the force exerted by the ball on the wall.
The force exerted by the ball on the wall is equal to the normal force N. Therefore, the ball pushes against the wall with a force equal to its weight:
Force(ball on wall) = N = mg
So, the ball pushes against the wall with a force of 45.0 kg (mass) multiplied by the acceleration due to gravity (9.8 m/s^2).
Where is the wire attached to the ball? If the wall contact point is frictionless, the wire attached to the ball must, if extended, pass thtough the center of the ball; otherwise the string will apply an unbalanced torque about the center.
The angle the string makes with the wall is given by
sin A = R/(R + 30) = 32/62
A = 31.1 degrees
Now apply a moment balance about the string contact point at the wall. The string tension applies no moment there since it passes through the point. If F is the contact force at the wall,
M g *32 = F * (30 + 32)cos 31.1
F = 45*9.8/[0.62 cos 31.1)= 830 N
You can get the string tension T from a hotizontal force balance.
F = T sin 31.1