Polygon of force diagram inclined plane

Click For Summary

Discussion Overview

The discussion revolves around the application of the polygon of forces to demonstrate the equilibrium of a crate on an inclined plane, specifically addressing the forces acting on the crate, including gravitational force components and normal force. Participants are exploring the correct representation of these forces in a diagram and the conditions for equilibrium.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the gravitational force (mg) and its components along the incline (fd) and perpendicular to the incline (Fn), but initially presents an incorrect value for Fn.
  • Another participant suggests recalculating Fn and proposes drawing the force components in a triangle, identifying mg as the hypotenuse.
  • A participant confirms the recalculated value for Fn and asks about the proper method for drawing the force triangle, considering scale and labeling.
  • There is a discussion about the conditions for equilibrium, with one participant noting that without friction, the crate cannot be in equilibrium as it would slide down the slope.
  • Another participant expresses confusion about the number of forces to consider, suggesting five forces, including friction and gravity, while others argue that only three forces are necessary for the analysis.
  • One participant mentions the role of friction in preventing the crate from slipping, while another clarifies that the gravity force is effectively split into components, which may explain its absence in some examples.
  • A separate participant asks about drawing forces at specific angles and magnitudes, indicating a need for further clarification on graphical representation.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the number of forces to consider in the analysis and the conditions for equilibrium, with some asserting that friction is necessary for equilibrium while others suggest it is not required in their specific context.

Contextual Notes

Some calculations and assumptions regarding the forces acting on the crate are subject to verification, and the discussion reflects varying interpretations of the problem setup, including the role of friction and the necessity of certain forces in the equilibrium analysis.

Who May Find This Useful

Students working on problems involving forces on inclined planes, particularly those learning about equilibrium and force diagrams in physics.

rad10k
Messages
60
Reaction score
0

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 don't understand please help! :o)
 
Physics news on Phys.org
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.
 
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? :o$

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

Thanks for your help
 
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:
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.
 
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
 
rad10k: Yes, very good. I like that.
 
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.
 
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
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
21
Views
4K
  • · Replies 13 ·
Replies
13
Views
6K