# Newton's 3rd Law issue

1. Nov 24, 2015

### Lisciu

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Hi Guys,

I was highly studied the Newton's law once again. Probably to recall the concept from the past of my education. Anyway I find out the exercise during solving the friction problems from Hibbeler book.

The exercises is state as :
The fork lift has a weight W1 and center of gravity at G. If the rear wheels are powered, whereas the front wheels are free to roll, determine the maximum number of crates, each of weight W2 that the fork lift can push forward. The coefficient of static friction between the wheels and the ground is μ s and between each crate and the ground is μ 's.

The whole system looks like this:

And then it's seperate on two objects:

FBD of the 1:

FBD of 2:

And there is nothing suprised about that except the action and reaction force from the crates and front wheel.

I expected that those force will be equal like the 3rd Newton law said. But during the solving and after see those images from book solution looks like 3rd Newton law doesn't work anymore....

Could someone explain what is the reason of that? And where is actually the action and reaction force? I'm confused here.

2. Nov 24, 2015

### Staff: Mentor

The only 3rd law pair here is $P=-P'$. If you got anything else then you made a mistake.

3. Nov 24, 2015

### Lisciu

Seriously? :) when you figured it out that I don't know that?

So why P is equal 707,37lb : and P' is equal : 105lb?

4. Nov 24, 2015

### Staff: Mentor

That is not correct, but there is no way to tell where you went wrong without more details about how you attempted to solve it.

5. Nov 24, 2015

### Lisciu

It's okay. I figured it out what is going on here. Because in the solving book author calculated the P' force for just one crate to calculate how many he can push by this fork lift. I realize that if I will calculated it with diffrent approach like typical 3rd law Newton without calculating reaction force for just one crate I will have the same result. I just took the calculate force P from fork lift and then using this force in X direction net force summing (2rd law) get the force of normal acting on. When I have it I just divided it by one crates weight and actually I get the number of crates this system can handle.

6. Nov 24, 2015

### Staff: Mentor

That's good. I think I probably would have followed something closer to your approach than to the books approach.

7. Nov 24, 2015

### Lisciu

Yeap, me too. I was just confused for whole day. Then I just sit once again on the problem and realize where was the hatch.

Anyway I will need to create the topic around 3rd Law because I not undestand in 100% this concept specially if someone talk about cancelling force not cancelling force and then i was thinking about this action and reaction force as a unended storry with domino effect in whole body if we consider for example hammer and nail.

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