Finding Force Applied on a ball from another object

  • Thread starter Drake M
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Homework Statement


Another question regarding pool. I need to know how much force applied is given from the cue to the ball as well as force of friction

OF CUE:
a=56.55m/s2
vi=0m/s
vf=3.28m/s
d=.0953m
m=.55kg
BALL:
m=.17kg
*not sure whether to use distance travelled during contact or of the ball after it was contacted as well as if to use vf of the cue as vi of the ball

Homework Equations


vf*2=Vi*2+2ad
W=Fd
Fnet=Fa+Ff
Co-efficient F= Ff x Fn

The Attempt at a Solution


For FNET of the ball i found acceleration using .02m as distance travelled during contact as well as vi=0m/s and vf=3.28m/s. Using vf*2=Vi*2+2ad I got an a of 268.75m/s2

Then using fnet=ma, I fot 45.65N as fnet of the ball. I also did the Ek=(1/2)mv*2 then w=fd to find fnet but results were the same.

Where I'm completely stuck is how to find the applied force on the ball.

I went and found fnet of the cue using Ek equation (m=.55kg, vf=3.28m/s) then used w=fd(d=.0953m) and got 30.95N. I felt good but then used FNET=fa+ff to get force of friction and it came out as 14.7N which wouldnt be weird if it werent for the fact that it is greater than the normal force so the coefficient of friction is larger than 1. I know that its possible but having that high of a coefficient doesnt make sense for a pool table.

Main goal is to find out what the applied force was and how much it is. ALSO, FORGOT TO MENTION IF THE FRICTION IS STATIC OR KINETIC. Thanks in advance to everyone
 

Answers and Replies

  • #2
haruspex
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First, 2cm is one hell of a contact distance for a pool shot.

Your calculation for the applied force seems to assume that the stroke causes the cue to come to rest, thereby using all of its KE. Would that be right? There was no tendency for the player to arrest the cue?

Using W=Fd to find F only really works if the force is constant. If it is not constant then it will not even give you the average force. But that's a minor point here.
 

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