How Do Forces Determine the Movement of a Block on an Inclined Plane?

[skateza]

Homework Statement

A block with a mass of 3.1kg is placed at rest on a surface inclined at an angle of 45 degrees above the horizontal. Th coefficient of static friction between the block and the surface is 0.50, and a force of magnitude F pushes upward on the block, parallel to the inclined surface.
a) The block will remain at rest only if F is greater than a minimum value Fmin, and less than a maximum value, Fmax. Explain the reasons for this behaviour
b) Calculate Fmin
c) Calculate Fmax

The Attempt at a Solution

Well i found a) to be IF you don't apply enough force (Fmin) the weight of the box will pull it down the surface, IF you apply too much force (Fmax) you will overcome the static friction of the box, pushing the box up the ramp.

My problems are in parts b and c;

I know that Fs is whatever it has to be in order to keep the object at rest until it reaches Fsmax. I also know once it is at max it is equal to UsN (0.5*mgcos45). So part c was really not to hard, just solve Fmax = mgsin45 - UsN, i got 10.75N

but in part b), if the box is not falling down the ramp than F >or= Fs, so i set Fs equal to F and solved the x component of the vector, 2F = mgsin45, i got 10.75N... How can i have too values that are identical for Fs and Fsmax... where did i go wrong?

 

[Kurdt]

So the force to overcome static friction you have calculated is $F_s=0.5\times mgcos(45)$. The component of weight acting parallel to the surface is $mgsin(45)$. So the minimum force required will be the force due to the weight acting parallel to the surface minus the force needed to overcome static friction.

Now do you think you can go on from there and say what the maximum force will be?

 

[m2003]

Your understanding of the concept is correct. However, in part b, you are only calculating the minimum value of F that is necessary for the block to remain at rest. This minimum value is equal to the static friction force (Fs) required to balance the weight of the block (mg) down the incline. This is why you are getting the same value for Fmin and Fmax.

To calculate Fmin, you can use the equation Fs = UsN = 0.5*mgcos45. This will give you a value of 3.9N. This means that as long as the applied force (F) is greater than 3.9N, the block will remain at rest.

Hope this helps clarify the concept for you!