# Forces coursework question

1. Jan 12, 2010

### black_hole

1. The problem statement, all variables and given/known data

A student pulls a box of books on a smooth horizontal floor with a force of 138 N in a direction of 23.9° above the horizontal. If the mass of the box and the books is 51.6 kg, what is the acceleration of the box?

2. Relevant equations

3. The attempt at a solution

I got a force diagram. I know that Fg = 505.68N and that should be equal to Fn. I know that Ft is the force causing acceleration. What do I do from there?

2. Jan 12, 2010

### Staff: Mentor

Re: Forces

Good.
Right.
Not right. That pulling force, since it's at an angle, will affect the normal force. Luckily, that doesn't matter for this problem since there's no friction.
What's the component of the pulling force in the direction of motion?
Apply Newton's 2nd law. What's the net force?

3. Jan 12, 2010

### black_hole

Re: Forces

Do you want to help me with this other one too?

A constant force of 2.2 N is required to drag a 49.2 kg box across a rough wooden floor at a constant speed of 2.2 m/s. Find the coefficient of sliding friction between the floor and the box.

I think Fn= Fg this time (b/c it is not at an angle), but what is Ff?

4. Jan 12, 2010

### Staff: Mentor

Re: Forces

Good.
Hint: Since the box moves at a constant speed, what's the net force on it? (That should allow you to solve for Ff.)

5. Jan 12, 2010

### MiniSmSm

Re: Forces

I dont understand !
the box moves at a constant speed
so the net force would be F = 0
but what about friction ? wouldnt be 0 ?
I`m confused

6. Jan 12, 2010

### Staff: Mentor

Re: Forces

That's true.
No, the friction isn't zero. (Friction is just one of the forces acting on the box.)

7. Jan 12, 2010

### black_hole

Re: Forces

So does that mean that Ff = 2.2 N? That would mean that the coefficient of friction would be 0.005?

8. Jan 12, 2010

### Staff: Mentor

Re: Forces

Right.
I would round off the answer to two significant figures, not one. But yes.

9. Jan 12, 2010

### black_hole

Re: Forces

A 7.4 kg box is released on a 30.6° incline and accelerates down the incline at 0.55 m/s2. What is the coefficient of kinetic friction between the two surfaces?

I found Fg = 72.52 N and Fgperpendicular = 62.421 N = Fn, Fgparallel = 36.916 N. But again I don't know how to find Ff?

10. Jan 12, 2010

### Staff: Mentor

Re: Forces

Hint: Use Newton's 2nd law to find the net force.

11. Jan 12, 2010

### black_hole

Re: Forces

I don't get it? Fgparallel - Ff = Fnet = 72.52 N. So, Ff = 35.604?

12. Jan 12, 2010

### Staff: Mentor

Re: Forces

That's correct.
That's incorrect. Fnet ≠ Fg. Use Newton's 2nd law to get Fnet.

13. Jan 12, 2010

### black_hole

Re: Forces

Thank you. You've been a big help!