# Determine the force to enable the object to move down

1. Nov 28, 2015

### werson tan

1. The problem statement, all variables and given/known data
A force P acts as shown on 200kg block placed on a inclined plane . The static coefficient between the block and the plane is 0.5 . Determine the force P to ensure the block is in the verge of moving down.

2. Relevant equations

3. The attempt at a solution
Wsin10 +P cos 20 = Fs
P cos 20 = 0.5( P sin20 - W cos 10 ) -Wsin 10
P cos 20 = 0.5( P sin20 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ?

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2. Nov 28, 2015

### azizlwl

20° angle is not reference to the inclined plane but to ground level.

3. Nov 28, 2015

### SteamKing

Staff Emeritus
For one thing, the force P is applied 20° above the horizontal. You must calculate the components of P relative to the block on the incline to determine the correct normal force of the block.

4. Nov 29, 2015

### werson tan

P cos 20 = 0.5( P sin20cos10 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ? my ANS STILL THE SAME

5. Nov 29, 2015

### azizlwl

It should be PCos(20 +10) not PCos(20)....and you to adjust that Sin 20 too.

6. Nov 29, 2015

### werson tan

THANKS , I NOTED MY MISTAKE , I GT THE ANS FINALLY