Determine the force to enable the object to move down

[werson tan]

Homework Statement

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.

Homework Equations

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 ?

 

[azizlwl]

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

 

[SteamKing]

Homework Statement

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.

Homework Equations

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 ?

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.

 

[werson tan]

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.

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

 

[azizlwl]

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

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

 

[werson tan]

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

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