Pushing a box across a horizontal floor

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Homework Help Overview

The problem involves determining the force required to push a 60 kg cardboard box along a horizontal floor, considering the effects of static and kinetic friction. The force is applied at an angle of 40 degrees above the horizontal, and the coefficient of static friction is given as 0.7.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to analyze the forces acting on the box, including the normal force and frictional forces. There are attempts to clarify the direction of the applied force and its components. Some participants express uncertainty about how to visualize the force's direction and its implications for the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the direction of the applied force, but there is still confusion among participants about the force's orientation and its effects on the box.

Contextual Notes

Participants note the absence of a diagram in the problem statement, which contributes to the confusion regarding the force's direction. There is also a focus on ensuring that the equations used reflect the correct understanding of the forces involved.

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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

 
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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.
 
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
 
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?
 
Benny851 said:
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.
 
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.
 
Hi Benny851! :smile:
Benny851 said:
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. :wink:

(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)
 

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