how come you set ma = umg?

[GomerPyle]

More questions by gomer 🙂

I encountered a question that doesn't explain this in the answer. The object was sliding down a ramp towards the ground, and at ground level it slid from point A to B where it comes to a stop by friction. In order to find the acceleration vector of the box, umg was set equal to ma. I got this question right because i knew to set umg to ma, but i dont understand why? So when you have a box that is sliding on the ground with kinetic friction acting opposite of motion, how come you set Fk = Fo (which becomes umg = ma where mass cancels and you get ug=a)?

*sorry, needs to be moved to questions/answers section.

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[Nickmiller]

When the box is sliding on a level ground, the Normal force is equal to the force of gravity. Since Force of gravity = mg, normal force (Fn) also equals mg. Force of friction (Ff) equals u (constant) times the normal force, so Ff=uFn=umg. If force of friction is the only force acting on the object, then Fnet=Ff, Fnet=umg, and since Fnet=ma (by definition), ma=umg.
Hope this explanation makes sense.

 

[balla121]

More questions by gomer 🙂

I encountered a question that doesn't explain this in the answer. The object was sliding down a ramp towards the ground, and at ground level it slid from point A to B where it comes to a stop by friction. In order to find the acceleration vector of the box, umg was set equal to ma. I got this question right because i knew to set umg to ma, but i dont understand why? So when you have a box that is sliding on the ground with kinetic friction acting opposite of motion, how come you set Fk = Fo (which becomes umg = ma where mass cancels and you get ug=a)?

*sorry, needs to be moved to questions/answers section.

So we have a box...(its helpful to draw it out).
From the top of the ramp to point A, there will be gravity component that is parallel to the ramp that "acts" (applies a force) on the box to accelerate the box; as it reaches point A, there will no longer be a force component from gravity and this is where the kinetic friction takes over. The kinetic friction will act against the motion of the box and the kinetic friction force will be the only force acting on the object(if object moving A to B is left to right, friction force is acting to the left). Whenever we have a net force, there will be an acceleration; in this case the net force is equal to the kinetic friction force. Thus, that is why we set umg(kinetic friction force) equal to ma. The acceleration in this case is negative(opposite of direction of motion)

 

[GomerPyle]

Thanks guys.

The only thing I don't understand is why we set the frictional force to ma. Is that just standard? The 'ma' is pointed to the left, in the same direction friction is, correct? I always assumed ma was in the direction of motion..

 

[3Pascals]

We set the frictional force to ma because the net force is always equal to ma (Newton's second law) and in this case the frictional force is the only force acting (technically the gravitational and normal forces are also acting but they cancel out).

Force need not be in the same direction as the motion of the object. Here, the object is moving to the right (velocity vector is pointed to the right) but the net force is pointed to the left and slowing the object down (acceleration is to the left). Think of it this way: the velocity vector points in the direction the object is moving; the force/acceleration vector points in the direction of the net "push" acting on the object.