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A sled weighing 60 N is pulled horizontally across snow so that the coefficient of kinetic friction between the sled and the snow is 0.100. A penguin weighing 70 N rides on the sled. If the coefficient of static friction between the penguin and the sled is 0.700, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off?

I got an answer of 49 N by just using the penguin's forces and the Ff=UsFN equation, but something tells me I need to use forces and the coefficient of kinetic friction for the sled as well. Any ideas?

You are right. If there were NO friction between the sled and snow, then the answer would be 49N. But, you also have to overcome the friction between the sled and the snow, which is?

Hmm.. I'm lost now. I tried finding the force applied to the sled but realized I didn't have an acceleration. So I just used the 49N I got from the penguin, thinking it was the same applied force. I got an acceleration of 7.02m/s^2 but then I don't know what to find and where to go from there. Or if what I did even makes sense in the first place/

OK. Think of it this way. You have a sled on the snow, and a penguin on the sled. You are applying a horizontal force so that the penguin will start to slide off. The penguin will only start to slide off when the force that's pulling on the sled overcomes the force of static friction between the penguin and the sled. And that force is 49N. But remember, you also have to add an additional force to get the sled to move over the snow in the first place, and that force is the force of friction between the sled and the snow. You don't need acceleration at all here.

But if the sled is already in motion and they told me the coefficient of kinetic friction, don't I need to have acceleration in my formula for the x-direction? -Ff+FA=ma where FA is the applied force and Ff is the force of friction. Sorry if I'm missing something blatantly obvious.

Armando, I am sorry, you are correct. I read the problem incorrectly that it was static friction between sled and snow!! Yikes! OK, lets start again and rethink this.

Yeah, I was wondering how acceleration didn't matter. Alright, so where did I go wrong then? I still don't think I did my work correctly :(

OK wait. Do we agree that we need to apply a NET force of 49N to the sled to overcome the static friction between the penguin and the sled?

Yes, that is agreed. The penguin will NOT move unless a force greater than 49 N is applied.

OK. So, your equation -Ff + FA = ma.

Can we not say that ma here is 49N?
The force of kinetic friction (the force that is opposing the horizontal force once it's in motion is 0.1(60N + 70N). So, Force applied is simply 49N plus this force of friction! Whadya think?

So you treat it as a whole in order to find the force of friction then? I guess that works, it makes sense. So the maximum force that you can safely apply is 62 N. Excellent. I'll check with my teacher tomorrow to make sure that's correct.

Thank you for the help.

Yes, I believe that's right!
If you get a chance, repost here tomorrow if your teacher disagrees! I'd be very curious!


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