Forces needed to slide up the plane

[goldfish9776]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[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?)

 

[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.

 

[haruspex]

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.

It's there, but with the wrong angle.

 

[goldfish9776]

It's there, but with the wrong angle.

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 ...

 

[haruspex]

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 ...

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

 

[goldfish9776]

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

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

 

[haruspex]

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

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?

 

[goldfish9776]

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?

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

 

[haruspex]

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

Isn't the angle still wrong?

 

[goldfish9776]

Isn't the angle still wrong?

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

 

[haruspex]

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

Ok!