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Find Force Magnitude Problem

  • Thread starter dgibbs
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  • #1
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Find Force Magnitude Problem...Please Help!

Homework Statement



1. You Push a block of mass 3.00 kg against a vertical wall by applying a force of magnitude P at an angle of 50.0 degrees above the horizontal. The coefficient of static friction between block and wall is 0.250. Determine the possible values of P that allow the block to remain stationary

Please help! I honestly am at a lost for what to do, can someone atleast start me off with the right equations? Thanks


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
798
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Draw the FBD of what is happening. You should also develop coordinate axes. What forces are acting on the object that is being held against the wall?
 
  • #3
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There is the Force with a magnitude P, that is unknown, that is pushing against the wall at the angle 50.0 degrees. = Fpsin(50.0)

There is a gravitational force of 9.8 m/s^2 : mg = 9.8m/s^2 x 3.00 kg =
29.4 N

There is a normal force of the block against the vertical wall = Fn
 
  • #4
798
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What about static friction?
 
  • #5
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And there is the Static Frictioin force between the block and the wall but all we need to know is the coefficent is 0.250
 
  • #6
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The sum of the forces in the Y direction is 0 right...

So ƩFy = 0

F(static friction coefficent) + Fpsinθ + -mg = 0
 
  • #7
798
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Okay. The force of friction is the value that is going to prevent the object from sliding. Note, however, the force of friction is a function of the normal force, which is provided by P.
 
  • #8
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Okay. The force of friction is the value that is going to prevent the object from sliding. Note, however, the force of friction is a function of the normal force, which is provided by P.
Okay, so is it...

Fμ(Fpsin(50.0)) + Fpsin(50.0) - mg = 0
(0.250)(Psin(50.0) + Psin(50.0) - (29.4N) = 0

Then solve for P?
 
  • #9
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The force P is acting horizontally at an angle of 50.0 degrees. The Force P seems to be merely holding the mass in place neither pushing it up or helping it slide down. It is possible that box could be sliding up or down depending on the magnitude of P.
 
  • #10
798
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The force P has a component acting in the same direction as the force of gravity (sine function). It also has a component acting in the direction of the normal force (cosine function).
 
  • #11
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The force P has a component acting in the same direction as the force of gravity (sine function). It also has a component acting in the direction of the normal force (cosine function).
Okay, so the force P must be negative to show its relationship with gravity and I guess I should also add Fpcos50.0 to the ƩF to represent the normal force?
 
  • #12
798
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Yes, that would be the normal force. Essentially, you will have two forces acting in the down direction, and one in the up direction.
 
  • #13
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Okay so to sum up:
Fmg = (-) because it is going downwards
Fpsin50.0 = (-) because it is going downwards
Fn = Fpcos50.0 **but is not (-)? or is also (-)**

So if these are the forces, I am still confused on one more topic...where does the friction coefficent come into play if Fn is a function of the normal force, do I just multiple the static friction coefficent with Fn.
 
  • #14
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Nevermind I think I understand the Fn part, Fn is the positive force.
 
  • #15
798
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[itex]F_f=\mu N[/itex]
N is the normal force. The force of friction will be directed up to oppose the sine component of P and the force of gravity.
 
  • #16
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Okay.
So we have accounted for all the Forces.
Ff=μN = (0.250)(Pcos50.0)
Fp = - Psin50.0
Fg = - mg(m) = -29.4N

Now I can isolate and solve for P...thank you very much Ms. Sandy Bridge, you were extremely helpful!!
Have a great evening
 
  • #17
798
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Good job!
 

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