Forces needed to slide up the plane

In summary: So the angle is wrong, and you need to correct it before solving.Ok! So the angle is wrong, and you need to correct it before solving.
  • #1
goldfish9776
310
1

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

 

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  • #2
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
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
andrewkirk said:
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.
 
  • #5
haruspex said:
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 ...
 

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  • #6
goldfish9776 said:
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.
 
  • #7
haruspex said:
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 ?
 
  • #8
goldfish9776 said:
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?
 
  • #9
haruspex said:
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
 
  • #10
goldfish9776 said:
P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 -100x9.81sin30 )
P=7331N
Isn't the angle still wrong?
 
  • #11
haruspex said:
Isn't the angle still wrong?
P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 -100x9.81sin20 )
P=7331N
 
  • #12
goldfish9776 said:
P cos30 +100x9.81xcos20 -1000x9.81xsin30 = 0.2 ( 1000x9.81cos30 +P sin30 -100x9.81sin20 )
P=7331N
Ok!
 

1. What is the definition of "forces needed to slide up the plane"?

The forces needed to slide up the plane refer to the amount of force required to move an object up an inclined surface.

2. What factors influence the forces needed to slide up the plane?

The forces needed to slide up the plane are influenced by several factors, such as the angle of the incline, the weight of the object, and the coefficient of friction between the object and the surface of the plane.

3. How do you calculate the forces needed to slide up the plane?

The forces needed to slide up the plane can be calculated using the formula F = mg sinθ, where F is the force needed, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.

4. What is the role of friction in determining the forces needed to slide up the plane?

Friction plays a crucial role in determining the forces needed to slide up the plane. The coefficient of friction between the object and the surface of the plane determines how much force is needed to overcome the resistance and move the object up the incline.

5. How can the forces needed to slide up the plane be reduced?

The forces needed to slide up the plane can be reduced by decreasing the angle of the incline, reducing the weight of the object, and increasing the coefficient of friction between the object and the surface of the plane. Additionally, using lubricants can also help reduce the forces needed to slide up the plane by reducing friction.

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