God! I have tried everything. How do you do this?

Question

There is a 2 kg block being pushed up a vertical wall at a constant velocity for a dist of 3 m and the coeff. of friction is .30. A constant force is applied at an angle of 28 deg with the horizontal. Find the magnitude of normal force btw the block and the wall.
Now I found the horizontal component to be 17.32 but I can't find the vertical component. I've done everything possible but can't solve this.

Answer 1

along x axis (sin 28=.47 cos 28=.88)
N=F*0.88
F=1.13*N

along y axis
U*N+20=.47*F
(.3/.47)*N+(20/.47)=1.13*N
solving...........
N=86.7 N
F=97.9N
friction=26.01

Answer 2

You need to be sure you are keeping horizontal and vertical straight. There are two forces: one due to gravity since the block is moving at constant velocity, one due to the friction with the vertical wall since the block is moving. The frictional force is the coef of friction times the "normal" force (that is force acting in the horizontal direction pushing the block against the wall). The total force acting at 28 degrees is equal to the sum of the two forces. The vertical force is equal to the force of gravity. You know the angle and magnitude. You can solve by trig to get the total force. You can express the normal force in terms of the total force (it is that component of the force pushing the block against the wall) by using trig. good luck

Answer 3

sorry, I'm not God, so you weren't addressing your question to me.

seems idiotic to push it up a vertical wall. just lift it in free space to avoid the wall friction -- a simple redesign.
if there's friction with the wall, that means all of the lifting force isn't vertical, so you're wasting some horizontally.

what's this constant force at 28 degrees. is it applied to the block?

Answer 4

The vertical component is : 0 )

Answer 5

OMG - you're asking ME??? HEE, HEE - good luck....