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Basic incline plane

  • Thread starter zaddyzad
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  • #1
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Homework Statement



A trunk of mass m = 1.300 kg is pushed a distance d = 124 cm up an incline with an angle of inclination theta = 28.0 ° by a constant horizontal force P = 425 N (see figure).
The coefficient of kinetic friction between the trunk and the incline is 0.130.

a) Calculate the work done on the trunk by the applied force P.
b) Calculate the work done on the trunk by the frictional force.
c) Calculate the work done on the trunk by the gravitational force.

http://imgur.com/ydGgOn7 http://imgur.com/ydGgOn7

My guess: (P has a component 28degrees above its horizontal, and another one pointing down perpendicular to the plane)

The Attempt at a Solution



A) shouldnt this be W = F*D.... (425/cos28)*1.24
b) shouldnt this just be W = FF * D.... ((1.3gcos28+425tan28)*.13)*1.24
c)W=Fs*D... 1.3gsin28*1.24

I got all the questions wrong... can anyone tell me where I went wrong ?
 

Answers and Replies

  • #2
199
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My guess: (P has a component 28degrees above its horizontal, and another one pointing down perpendicular to the plane)

A) shouldnt this be W = F*D.... (425/cos28)*1.24
b) shouldnt this just be W = FF * D.... ((1.3gcos28+425tan28)*.13)*1.24
c)W=Fs*D... 1.3gsin28*1.24

I got all the questions wrong... can anyone tell me where I went wrong ?
For A), you should calculate force P along the incline, which is (425*cos28) not (425/cos28).

For B), you never showed any calculation for normal force!!!.....Calculate Normal Force again!!!!

For C), you did almost everything right except the sign. Gravity in this case is against the direction of motion.
 
  • #3
Simon Bridge
Science Advisor
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Work done by a force is equal to the component of the force in the direction of the motion, multiplied by the distance traveled.

Check the signs, the relations for the components of P, and the energy balance.

The energy is supplied by the work done by force P.
Some energy is lost to friction.
The rest goes into gravitational potential energy.
 
  • #4
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For A), you should calculate force P along the incline, which is (425*cos28) not (425/cos28).

For B), you never showed any calculation for normal force!!!.....Calculate Normal Force again!!!!

For C), you did almost everything right except the sign. Gravity in this case is against the direction of motion.
I'm solving for the hypotenuse, the parallel component to distance, what you just said would give me the adjacent side which is the side with the known value.
 
  • #5
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I'm solving for the hypotenuse, the parallel component to distance, what you just said would give me the adjacent side which is the side with the known value.
you seem to have some fundamental doubt. draw the force vector and its components, in form of a "triangle sum". you'll see what I mean.
 
  • #6
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I just got B and C, so now my focus is A.
 
  • #7
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I just got B and C, so now my focus is A.
great!!!
 
  • #8
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  • #9
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http://imgur.com/hBSSssV, I don't see where my mistake is.
You drew the triangle wrong. Force vector is the hypotenuse of the triangle, not simply its side.

Edit: If it still doesn't ring any bells, you'll need to check out vector sum and vector resolution.
 
  • #10
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425 N is the horizontal force parallel to the base of the incline.
 
  • #11
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h is what I solved for, and it is the force parallel to the direction of motion. I don't see my mistake anywhere..
 
  • #12
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425 N is the horizontal force parallel to the base of the incline.
Yes, it is. If you meant about my previous post, it was regarding the orange+black vector triangle.
 
  • #13
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I don't see what you're saying.
 
  • #14
Simon Bridge
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@zaddyzad: do you know how to find the components of a vector along x and y axis?

Draw the x axis so it points up the slope - in the direction of movement.
Draw the y axis so it points upwards away from the slope.
Now find the components of P.

The vector P should be the hypotenuse of the triangle made by the components.
The vector components are always smaller than the vector.
 
  • #15
199
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h is what I solved for, and it is the force parallel to the direction of motion. I don't see my mistake anywhere..
I don't see what you're saying.
Look at the figure attached. It will make things clear. Here z represents your force vector P. Ignore the equation written, I copied the image off the web.
 

Attachments

  • #16
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Ok, now that I got that. This now leaves me with the some new questions.

What was wrong with my vector sum diagram, I still had a component of force parallel to the plane?
And how did I get the right answer for by when I solved for the perpendicular force on the plane to be (1.3g + 425tan28) * cos28
 
  • #17
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I'm confused if I could calculate the correct vertical force with my "incorrect" triangle, why wouldn't it allow me to compute the correct parallel force?
 
  • #18
199
15
Ok, now that I got that. This now leaves me with the some new questions.

What was wrong with my vector sum diagram, I still had a component of force parallel to the plane?
And how did I get the right answer for by when I solved for the perpendicular force on the plane to be (1.3g + 425tan28) * cos28
Whenever you resolve a vector, that vector is always assumed to a hypotenuse of vector triangle. This assumption is correct. Simple trigonometry can prove it. The vector sum of adjacent sides always gives the original vector (hypotenuse).

If you check your method, then you won't the sum of new vectors to be same as the old one.
 
  • #19
199
15
I'm confused if I could calculate the correct vertical force with my "incorrect" triangle, why wouldn't it allow me to compute the correct parallel force?
Your can try, but it won't give you the right answer!!!......It also depends what do you mean by vertical, the y axis perpendicular to the incline or the regular vertical
 

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