Pushing a box across a horizontal floor

[Benny851]

Homework Statement

A man wants to push a cardboard box 60 kg along a floor by applying force F in a direction 40 degrees above horizontal. Coefficient of static friction between box and floor is .7.

a) Determine F if box is on verge of moving.

b) Based on part A, determine acceleration along the floor if coefficient of kinetic friction is 0.32.

Homework Equations

The Attempt at a Solution

 

[Nugatory]

C'mon, you have to at least try yourself before you ask for help...

If you're having trouble deciding where to start... Start by drawing a diagram of all the forces acting on the block. Write each one as a sum of a horizontal and vertical component.

 

[Benny851]

The problem is I don't know how to draw the force acting on the box. Is the force coming up from ground or going down onto the box from top left? No picture was supplied with problem

 

[Benny851]

OK, so please tell me if I am heading in right direction. The equations I have so far are as follows:

1. N = Fsin theta + MG
2. Fcos theta = MsN

Now all I need to do is solve for F and that will be my answer to part A, right?

 

[haruspex]

1. N = Fsin theta + MG

It says 'above horizontal', i.e. partly upwards, not downwards.

2. Fcos theta = MsN

Now all I need to do is solve for F and that will be my answer to part A, right?

Yes.

 

[Benny851]

which direction is the force acting on the box? I don't get what 'above horizontal means'? I assume the force is pointed downward at the top left corner of the box.

 

[tiny-tim]

Hi Benny851!

which direction is the force acting on the box? I don't get what 'above horizontal means'? I assume the force is pointed downward at the top left corner of the box.

No, "above" means the way the force is going to, not coming from …

so the man is pushing upwards against the back of the box at 40° above horizontal.

(so far as sliding is concerned, the effect should he the same as the more usual case of a rope attached to the front of the box and being pulled at 40° above horizontal)