Is the Upward Force on a Bouncing Ball the Same as the Average Normal Force?

  • Thread starter Thread starter issacnewton
  • Start date Start date
  • Tags Tags
    Ball Bouncing ball
AI Thread Summary
The discussion centers on the forces acting on a bouncing ball, specifically the relationship between the upward force exerted by the ground and the average normal force. It is established that the upward force is indeed the normal force at all times during contact with the ground. When the ball is in contact with the floor, two forces are present: the downward gravitational force (mg) and the upward normal force (N). The normal force changes due to the ball's momentum during the bounce. Overall, the free body diagram of the ball while in contact with the ground illustrates these two opposing forces clearly.
issacnewton
Messages
1,035
Reaction score
37
Hello

I just have some basic question about the bouncing ball. When we throw a ball on the gound,
there is usual force of mg downwards, but since the ball is bouncing, due to the change in momentum, the ground will exert an upward force on the ball. Is this upward force same as the average normal force when the ball is in contact with the ground ? I am trying to think of free body diagram of the ball while it is in contact with the ground before it bounces off.

thanks
 
Physics news on Phys.org
The upward force IS the normal force at all times
 
so how would the free body diagram look like ? when the ball is in contact with the floor, we have mg downwards and the N upwards, which is force exerted by the floor on the ball because of change in momentum ? so we have just two forces acting on the ball , right ?
 
IssacNewton said:
so how would the free body diagram look like ? when the ball is in contact with the floor, we have mg downwards and the N upwards, which is force exerted by the floor on the ball because of change in momentum ? so we have just two forces acting on the ball , right ?

yes, there are only 2 forces on the ball.
 
I appreciate willem2.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top