Coefficient of static friction of a sliding box

[DB91]

Homework Statement

A box rest on an incline making a 34 angle with the horizontal. It is found that a parallel force to the incline of at least 235 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.

Homework Equations

Just wanted to know what formula to use or at least how construct the formula.

The Attempt at a Solution

Didn't know where to start.

 

[NascentOxygen]

Homework Statement

A box rest on an incline making a 34 angle with the horizontal. It is found that a parallel force to the incline of at least 235 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.

Homework Equations

Just wanted to know what formula to use or at least how construct the formula.

Hi DB91. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You are not needing a formula yet! The first step is to draw a clear, neat diagram, to mark on everything you know here.

Post that when you've done it.

 

[DB91]

Hi DB91. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You are not needing a formula yet! The first step is to draw a clear, neat diagram, to mark on everything you know here.

Post that when you've done it.

https://mail.google.com/mail/u/0/?ui=2&ik=d83d1b3b4d&view=att&th=144c131a0bf54f59&attid=0.1&disp=emb&realattid=ii_144c1315a9dabb9a&zw&atsh=1

 

[DB91]

Would it be: (800×9.8 sin⁡34)/235 = f

 

[NascentOxygen]

https://mail.google.com/mail/u/0/?ui=2&ik=d83d1b3b4d&view=att&th=144c131a0bf54f59&attid=0.1&disp=emb&realattid=ii_144c1315a9dabb9a&zw&atsh=1

Your image doesn't show up. (I doubt that that is a link to an image, anyway.)

 

[DB91]

I'll try a different address

 

[NascentOxygen]

Your image is accessible now. (I think you have included a lot of whitespace in it?)

The question refers to an arrangement of forces needed "to prevent the box from sliding down" the incline. So this tells you the direction in which friction will be acting in your forces diagram. So, in what direction will friction be acting here, and why have you decided it must be in this direction?

 

[DB91]

Your image is accessible now. (I think you have included a lot of whitespace in it?)

The question refers to an arrangement of forces needed "to prevent the box from sliding down" the incline. So this tells you the direction in which friction will be acting in your forces diagram. So, in what direction will friction be acting here, and why have you decided it must be in this direction?

The friction force would be acting in the opposite direction of the force applied on the box. Right?

 

[NascentOxygen]

The friction force would be acting in the opposite direction of the force applied on the box. Right?

Not necessarily. It depends on what the applied force is tending to do.

How could you work out the correct answer to this?

 

[DB91]

Not necessarily. It depends on what the applied force is tending to do.

How could you work out the correct answer to this?

(800×9.8 sin⁡34)/235 = f...?

Sorry physics is defiantly not one of my strong points.

 

[NascentOxygen]

Well, if the box were sitting stationary on the incline, and needed no extra external force to keep it steady, what direction would you be drawing the arrow showing the friction force that is acting on the block?