# Homework Help: Forces needed to slide up the plane

1. Oct 30, 2015

### goldfish9776

1. The problem statement, all variables and given/known data
Pco20- Wsin30-100(9.81)(cos20) –Fs = 0

Pcos20 - 1000x9.81xsin30 – 100x9.81xcos20 – 0.2 x ( Wcos20x9.81 + Psin20 ) = 0

Pcos20 – 4905-921.8-1700-0.07P=0

7526.8= 0.869P

P= 8661N
what's wrong with my working ? the ans given is P=7330N

2. Relevant equations

3. The attempt at a solution

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2. Oct 30, 2015

### haruspex

At least two issues there.
Some of your 20 degrees should be 30 degrees.
You have a sign wrong. (Which way does the tension act?)

3. Oct 30, 2015

### andrewkirk

There should be a term $1000\times g\times\cos 30^\circ\times \mu_s$ in there somewhere for the frictional force from the weight of the 1000kg block on the ramp. I can't see any such term in your calc.

4. Oct 30, 2015

### haruspex

It's there, but with the wrong angle.

5. Oct 30, 2015

### goldfish9776

the block is in the verge of moving up , so the tension of rope should look like this ?
P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 )
P=7419N ....

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6. Oct 30, 2015

### haruspex

Much closer, but you've left out a contributor to the normal force.

7. Oct 31, 2015

### goldfish9776

P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 + 100x9.81sin30 )
P=7506N ????

8. Oct 31, 2015

### haruspex

A couple of problems with the term you added.
Which way does this force act, in terms of the total normal force?
What angle does it make to the normal?

9. Oct 31, 2015

### goldfish9776

P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 -100x9.81sin30 )
P=7331N

10. Oct 31, 2015

### haruspex

Isn't the angle still wrong?

11. Oct 31, 2015

### goldfish9776

P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 -100x9.81sin20 )
P=7331N

12. Oct 31, 2015

### haruspex

Ok!

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