Polygon of force diagram inclined plane

  • Thread starter rad10k
  • Start date
  • #1
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Homework Statement



Use the polygon of forces to demonstrate that the crate is in equilibrium on the slope

Homework Equations



No friction 1962 N crate sitting on a inclined plane 25 degree angle.

The Attempt at a Solution



I have drawn four forces

mg = 1962 N ( 200 kg * 9.81)

fd = mgsin25 = 829.14 N

Fn = mgcos25 = 1944.74 N

So I have resolved the component forces how do I now draw it to demostrate that the crate is in equilibrium and the polygon or right angled triangle will "close "

I dont understand please help!!! :eek:)
 

Answers and Replies

  • #2
nvn
Science Advisor
Homework Helper
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rad10k: Your numeric value for Fn is currently incorrect. Try computing that again.

You can draw those two force components in a force triangle, where m*g is the hypotenuse.
 
  • #3
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thanks nvn . I have recalculated fn and now have fn = mgcos25 = 1778.18 N

Do I have to draw the triangle to scal ie. 1 cm = 100 N or do I just draw a triangle and label it with the units of force with it? :eek:$

mg as the hypotense , fd as the opposite fn as adjacent?

Thanks for your help
 
  • #4
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I have just drawn this on a scale of 1 cm to 100 N and the mg hypotense side measured out correctly , also I checked angle with a petractor and the angle is 25 degrees so is this now correct? thanks
 
Last edited:
  • #5
nvn
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rad10k: Yes, that sounds correct. Nice work. Just a footnote: The crate will slide down the slope unless there is a frictional force equal to fd. If the crate slides (accelerates), it is not in equilibrium. With no friction, the crate cannot be in equilibrium.
 
  • #6
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thanks nvn .

I have added a free body diagram with force fk in oppositie direction to fn which i take it will also be 829.18 N and also added force n "normal force" which I believe is equal to fn? which 1778.18 N

Anyway thanks for your time
 
  • #7
nvn
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rad10k: Yes, very good. I like that.
 
  • #8
Hi guys,

Sorry for jumping in on this but I'm a little confused too! I have the same problem. 200kg crate on a 25 degree plane.

I'm confused with the force diagram. I thought I was looking to balance 5 forces?

1) The component of the weight acting parallel to the slope.
2) The component of the weight acting perpendicular to the slope.
3) The friction, acting parallel to the slope.
4) The reaction force of the slope on the block.
5) The gravity force through the centre of mass.

Looking at other examples the gravity force isn't used. Is this due to the force being split into the vertical and horizontal components?

Thanks in advance for any help.

Rich.
 
  • #9
To solve this you have to consider only 3 forces

the weight, the normal reaction force, the force pulling the crate up the hill.

the role played by friction is just to keep the drate from slipping.
 
  • #10
Hi, how do I draw 90N;90degrees 150N;60degrees and 75N;75degrees and 100N
 

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