Minimizing Pulling Force for Crate on Rough Floor

  • Thread starter Thread starter ethex
  • Start date Start date
  • Tags Tags
    Friction
Click For Summary

Homework Help Overview

The problem involves determining the angle at which a person should pull a crate across a rough floor to minimize the pulling force while maintaining constant velocity. The crate has a mass of 100 kg, and the coefficient of kinetic friction is 0.4.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the crate, including friction and the normal force. There are attempts to derive equations relating the pulling force and angle, with some questioning the role of the normal force and the assumptions made about the forces involved.

Discussion Status

Participants are actively engaging with the problem, with some providing equations and others questioning the assumptions and definitions used in the setup. There is no explicit consensus, but there are suggestions for further exploration, such as drawing a diagram to clarify the forces involved.

Contextual Notes

Participants note the importance of including all forces in the analysis, particularly the normal reaction force, which is affected by the angle of the pulling force. There is an emphasis on ensuring that the equations account for the correct relationships between the forces.

ethex
Messages
25
Reaction score
0

Homework Statement


A person is dragging a packing crate of mass 100kg across a rough floor where the coefficient of kinetic friction is 0.4. He exerts a force F just sufficient to keep the crate moving at a constant velocity. At what angle above the horizontal should his pulling force F be for it to be minimum?

Homework Equations





The Attempt at a Solution



Kinetic friction force = 0.4 x 100g = 40g

Fcos theta = 40g
F sin theta = 100g

tan theta = 10/4
theta = 68.1 degree

am i right?
 
Physics news on Phys.org
Draw a diagram and put in ALL the forces. You are leaving out the normal reaction force of floor on crate. This normal reaction force IS NOT EQUAL to the weight of the crate since the pulling force is not horizontal.
 
Ok.

Let the angle be theta,

Friction force = 0.4 x 100g x cos theta

F - 100g sin theta - 40g cos theta = 0

F = 980 sin theta + 392 cos theta

dF/dtheta = 980 cos theta - 392 sin theta

Let derivative = 0

tan theta = 2/5

i got back the same ans! What is wrong?
 
ethex said:
Ok.

Let the angle be theta,

Friction force = 0.4 x 100g x cos theta

Why? What does cos(\theta) have to do with the weight of the crate?

What's the normal force?
 
The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.
 
The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
6
Views
3K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
5K