Determine the force to enable the object to move down

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Homework Help Overview

The problem involves determining the force required to move a 200kg block down an inclined plane, considering the static coefficient of friction between the block and the plane. The angles involved in the force application and the incline are crucial to the calculations.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the components of the applied force P and how they relate to the block's motion down the incline. There are attempts to set up equations based on forces acting on the block, including weight and friction.

Discussion Status

Several participants have pointed out potential errors in the calculations and assumptions regarding the angles involved. There is ongoing exploration of how to correctly resolve the components of the force P relative to the incline. Some participants have noted discrepancies between their calculations and the expected answer.

Contextual Notes

There is a mention of the angle of force application being above the horizontal, which may affect the normal force calculation. The discussion reflects a lack of consensus on the correct approach to resolving the forces involved.

werson tan
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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 ?
 

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20° angle is not reference to the inclined plane but to ground level.
 
werson tan said:

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.
 
SteamKing said:
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
 
werson tan said:
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.
 
azizlwl said:
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
 

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