More friction-related wackiness

[legking]

*ahem*

A box with a mass of 22 kg is at rest on a ramp inclined at 45 deg to the horizontal. The coefficients of friction between the box and the ramp are $$\mu_S = 0.78$$ and $$\mu_K = 0.65$$.

a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.

b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

Answers:

a) K, I have a FBD with an applied force acting upwards parallel to the ramp, static friction acting in the opposite direction, gravity acting directly downwards, and normal force acting away from, and perpendicular to, the ramp.

I have also designated the directions upwards both parallel to and perpendicular to the ramp as positive.

If the box is to remain at rest, the applied force must equal the friction and the portion of the gravity acting against the ramp, or Fa = Ff + Fgr.

(I also noticed that, at an angle of 45 deg, the vectors for Fg, Fn and Fgr form an isoscles triangle, making Fn equal to Fgr.)

So,

Fa = Ff + Fgr
Fa = (mu)S * mg cos 45 + mg sin 45
Fa = (0.78)(22)(9.8) cos 45 + (22)(9.8) sin 45
Fa = 118.95 +152.5
Fa = 271.5 N

b) So this is where I get a little lost. My FBD has applied force into the box perpendicular to the ramp, both static friction and the normal force acting in the opposite direction, and gravity acting downwards. I just don't know what to do to develop an equation that doesn't look exactly like the one I developed in a). Any suggestions?

 

[Doc Al]

a) K, I have a FBD with an applied force acting upwards parallel to the ramp, static friction acting in the opposite direction, gravity acting directly downwards, and normal force acting away from, and perpendicular to, the ramp.

I have also designated the directions upwards both parallel to and perpendicular to the ramp as positive.

If the box is to remain at rest, the applied force must equal the friction and the portion of the gravity acting against the ramp, or Fa = Ff + Fgr.

(I also noticed that, at an angle of 45 deg, the vectors for Fg, Fn and Fgr form an isoscles triangle, making Fn equal to Fgr.)

So,

Fa = Ff + Fgr
Fa = (mu)S * mg cos 45 + mg sin 45
Fa = (0.78)(22)(9.8) cos 45 + (22)(9.8) sin 45
Fa = 118.95 +152.5
Fa = 271.5 N

I didn't check your arithmetic, but your method is correct.

b) So this is where I get a little lost. My FBD has applied force into the box perpendicular to the ramp, both static friction and the normal force acting in the opposite direction, and gravity acting downwards. I just don't know what to do to develop an equation that doesn't look exactly like the one I developed in a).

There are several things different here:
(1) there are only two forces parallel to the ramp (the applied force is perpendicular to the ramp)
(2) the normal force is no longer simply mg cos 45.

Set up your equilibrium equation and solve for the applied force.