Kinetic Friction

[soby10]

---

Members do not see this ad.

From TPR Physics

A person is pulling a block of mass m w/ a force equal to its weight directed 30 degrees above the horizontal plane across a rough surface generating a friction f on the block. If the person is now pushing downward on the block with the same force 30 degrees above the horizontal plane across the same rough surface what is the friction on the block? Answer is 3f

Explanation states F=ukN where N is normal force acting on block. I understand that however it then states when force is applied 30 degrees above horizontal N=mg-mgsin30? How is that the normal force I thought the normal was mgcos30? It the states when force is applied 30 degrees below horizontal N=mg+mgsin30? Solving for N and plugging it into F=ukN and solving for uk gives 3f.

 

[Baller MD]

It's mgsin, brah.

 

[attixx]

I believe it has to do with Normal force.

Normal force can be equated with "tightness" between objects. If you are pulling up 30 degrees above the horizontal, you are decreasing it's effective weight. If you push down with the same angle, you are increasing the tightness between the block and the surface and thus the normal force. (and as I'm sure you are aware, normal force affects the force of friction.)
Consider a block sitting on a surface - the normal force simply counteracts the force of gravity (which is equal to mg). Now you are pulling 30 degrees to the horizontal - you are pulling with a certain amount of force in the opposite direction of the force of gravity. The normal force (the force counteracting the force in the downward direction] now equals [mg - mg(sin30)].

Normal force = [mg - mg(sin30)] in the case of pulling up. You are decreasing the normal force from its value of mg by decreasing the force of contact between the block and the surface.
Sin30 = 0.5. So for example, lets say m=1kg. The normal force is then [mg-mg(sin30)] which then becomes [(1kg)(10m/s^2)-(1kg)(10m/s^2)(1/2)] and when you plug in the variables ---> [10-5] = 5 N


The way I think of it is when you push in the opposite direction, the angle becomes (sin330) which equals -0.5. You are adding to the normal force. Remember, normal force simply counteracts the force contact force in the downward direction. Plug into the formula with the new negative value. Normal force = [mg - mg(sin330)]; Sin330 = -0.5.
Plug in the variables: [(1kg)(10m/s^2)-(1kg)(10m/s^2)(-1/2)] ---> becomes [10-(-5)] = 15N

Since the coefficient of friction is the same in both cases, the only difference is the normal force which differs by a factor of 3.

A little hard to explain on the forum, but I hope it made a little sense.

 

[soby10]

Thanks for the clear explanation. I thought the block was on an incline thus couldn't make out the answer.