Finding Friction Coefficient to Stop Big Box: Explained

[TSN79]

Below is a problem I had on a test recently. The task is to find out what the friction coefficient must be in order to prevent the big box from moving. Friction between the little box and the slope is neglected.The red markings are my ideas. I figured if I sum the forces in the x-direction I'll find the friction force, which I can replace by uN, since F=uN. Will N be the same as G? If so, finding u is easy...but I think it's more complex than that, can someone explain?

 

[Hootenanny]

I cannot see the image however, there is one thing the I always to when solving these type of problems: Resolve all forces so they are either parallel or perpendicular to the inclined plane. Then just sum the forces.

You say in your post;

Will N be the same as G?

The normal reaction force will only be equal to mg if the plane is horizontal. Otherwise you must resolve the forces to find N (which you should have done if you follow my advice above), the normal reaction force is always perpendicular to the surface.

Hope this helps.

~H

 

[TSN79]

N is perpendicular to the box, or in other words, parallel to G, since the box simply rests on flat ground, but there is a force pushing on the box sideways, so I'm just wondering if that will have an effect on N? I'm thinking and hoping not...

 

[Hootenanny]

N is perpendicular to the box, or in other words, parallel to G, since the box simply rests on flat ground, but there is a force pushing on the box sideways, so I'm just wondering if that will have an effect on N? I'm thinking and hoping not...

Then no, if the applied force is perpendicular to g, then it will have no effect on the normal reaction force of the surface on the block.

I'm afraid there's no mentors online at the moment that I can ask to approve your attachment.

~H