Simple physics stuff

[Lucifers Mentor]

Right - I'm trying to go over my old physics stuff, and I'm already running into troubles.

If you have equal forces pulling on an rope in opposite directions, why is the force on the center of the rope not equal to the some of the forces? I thought it would just be T - F1 = F2

Since I'm begging for help, reaction forces. If you look at a ball on bat situation, for the ball to rebound, wouldn't the force (on the ball by the bat) on impact have to be greater than the force on the bat by the ball? Because otherwise, how does the force exerted through a cricket bat (for example) through the stroke affect the motion of the ball - if the forces are equal and opposite the whole time, why do you need to swing?

Right, I'm already feeling stupid, so I'll leave it there

Thanks everyone

[Shaetano]

With the rope, its in equilibrium if the forces on both sides are equal (sum of the forces = 0). Now if you cut the rope in the middle the force in the middle (tension?) is what you need for it to remain in balance, so just F1.

For the second one, it's because you are forgetting Newton's second law (Force = Mass x Accel).

The reason you and the bat don't move in the opposite direction is that the force isn't big enough to overcome the friction you have with the ground (iirc anyway :P) (conservation of momentum maybe ?)


Hopefully thats correct, haven't had to use physics in a long time lol

[Goth]

A ball bounces because energy is stored up in compressing the ball, which lets it rebound. That, and some of the kinetic energy the ball originally has is present in the system still.

For example, a steel ball has a far smaller coefficient of restitution than a rubber ball.

[ChesterMcCheese]

Originally Posted by Lucifers Mentor

If you have equal forces pulling on an rope in opposite directions, why is the force on the center of the rope not equal to the some of the forces? I thought it would just be T - F1 = F2

it isn't?

Originally Posted by Lucifers Mentor

Since I'm begging for help, reaction forces. If you look at a ball on bat situation, for the ball to rebound, wouldn't the force (on the ball by the bat) on impact have to be greater than the force on the bat by the ball? Because otherwise, how does the force exerted through a cricket bat (for example) through the stroke affect the motion of the ball - if the forces are equal and opposite the whole time, why do you need to swing?

on impact, the force of bat on ball = force on ball on bat (newton's third law). of course if u swing the bat harder the ball gets hit harder. think about how fast the ball is moving relative to the bat. if the ball is travelling 10 and the bat is travelling 10 towards it, that's the same as the ball hitting a still bat at 20.

sorry if i misunderstood u

[fR33z3]

Originally Posted by ChesterMcCheese

if the ball is travelling 10 and the bat is travelling 10 towards it, that's the same as the ball hitting a still bat at 20.

That is a pretty simplified view of the world, but for very 1-dimensional newtonian physics it holds true.

A more precise way of saying this is that momentum is conserved in a "perfect" collision, where momentum is defined as the abs(v)*m.

If you have equal forces pulling on an rope in opposite directions, why is the force on the center of the rope not equal to the some of the forces

You are just thinking 1-dimensional (in fact, you are thinking 0-dimensional). You need to think vectors, not scalars. Force has direction! YOu are right....it is the sum of all forces, but to be more clear, it is the VECTOR sum of all forces.

Thinking 1-dimensional, a unit of force up, and a unit of force down, acting on the same object, will have a net force AS APPLIED TO THE OBJECT OF INTEREST of zero.

To reinforce this idea, by newton's 2nd law, F=ma, the object of interest will not be accelerated by a net force of zero. Assuming the rope does not have an initial velocity, the rope will remain unmoved.

This of course assumes no other external forces (such as gravity) and that the rope has no elastic properties. It should also be noted that the force in this scenario is applied to "the rope", not "the centre of the rope". However applying the above assumptions would yield the same answer if we were to consider an imaginary point anywhere along the rope.