# Finding acceleration of a block on an incline without friction, then with friction!

1. Oct 1, 2012

### riseofphoenix

7. A block of mass m = 5.5 kg is pulled up a θ = 22° incline as in the figure with a force of magnitude F = 38 N.

(a) Find the acceleration of the block if the incline is frictionless

So I got this one right, and here is what I did...

From the free body diagram I drew:

Fnet = (38 N) + (-mg sin 22)
ma = 38 - mg sin 22
a = (38 - mg sin 22)/m
a = [ 38 - (5.5)(9.81)(sin 22) ] / (5.5)
a = (38 - 20.211) / 5.5
a = 3.24 m/s2

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.

So...I know that

fk = μkn
fk = 0.12(Fappliedsin θ - mg)

QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = Fappliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better??? Because I almost got the answer...

fk = 0.12[38 sin 22 - (5.5)(9.81)]
fk = -4.767 (the negative number makes sense since kinetic friction is OPPOSING the motion of the block which is moving up the incline.

So then I add it to the equation for acceleration at the top

ma = (38 N) + (-mg sin 22) + (-4.767)
a = [ 38 - (5.5)(9.81)(sin 22) - 4.767 ] / (5.5)
a = 13.022 / 5.5
a = 2.36 m/s2

This answer is wrong...Webassign says the answer is actually a = 2.15 m/s2

QUESTION 2: Where did I go wrong? Was it the normal force equation part?

2. Oct 1, 2012

### Staff: Mentor

Re: Finding acceleration of a block on an incline without friction, then with frictio

Right.
Not right.
The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

3. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

No wait...
it would be

-mg + (-mg sin 22)?

4. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

It's just F cos θ, which is 38 cos 22...

But I still don't get the right answer when I multiply 0.12 by 38 cos 22, and then add 4.22 to this equation...

ma = 38 - 20.11

a = (38 - 20.11 + 4.22) / 5.5 doesn't give me 2.15 m/s2

Something else is wrong

Last edited: Oct 1, 2012
5. Oct 1, 2012

### Staff: Mentor

Re: Finding acceleration of a block on an incline without friction, then with frictio

The applied force F is parallel to the incline. It has no normal component.

The weight of the block, being vertical, will have a normal component.

6. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

So it's -mg cos 22?
Which is basically -(5.5)(9.81)(cos 22)...

fk = 0.12[-(5.5)(9.81)(cos 22)]
fk = -6.003

So,

a = [ 38 - 20.211 + (-6.003) ] / 5.5
a = 2.159638636

:)

Thanks!

7. Oct 1, 2012

### Staff: Mentor

Re: Finding acceleration of a block on an incline without friction, then with frictio

Almost. What's the weight of the block?

8. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

It's just 5.5 kg isn't it?

9. Oct 1, 2012

### Staff: Mentor

Re: Finding acceleration of a block on an incline without friction, then with frictio

No, that's the mass.

10. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

oh...Weight = mg
Weight = -(5.5)(9.81) = -53.955 N

11. Oct 1, 2012

### Staff: Mentor

Re: Finding acceleration of a block on an incline without friction, then with frictio

Right. The weight = mg and acts downward.

12. Oct 1, 2012

### riseofphoenix

Re: Finding acceleration of a block on an incline without friction, then with frictio

Ohh ok!
Thanks a lot!!!!