How Do You Calculate the Acceleration of a Box on a Smooth Floor?

[black_hole]

Homework Statement

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?

Homework Equations

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?

 

[Doc Al]

I got a force diagram.

Good.

I know that Fg = 505.68N

Right.

and that should be equal to Fn.

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.

I know that Ft is the force causing acceleration.

What's the component of the pulling force in the direction of motion?

What do I do from there?

Apply Newton's 2nd law. What's the net force?

 

[black_hole]

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?

 

[Doc Al]

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

Good.

but what is Ff?

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

 

[MiniSmSm]

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

 

[Doc Al]

the box moves at a constant speed
so the net force would be F = 0

That's true.

but what about friction ? wouldn`t be 0 ?

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

 

[black_hole]

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

 

[Doc Al]

So does that mean that Ff = 2.2 N?

Right.

That would mean that the coefficient of friction would be 0.005?

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

 

[black_hole]

Now I know I'm being really annoying, but what about this one...

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?

 

[Doc Al]

But again I don't know how to find Ff?

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

 

[black_hole]

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

 

[Doc Al]

Fgparallel - Ff = Fnet

That's correct.

= 72.52 N.

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

 

[black_hole]

Thank you. You've been a big help!