Forces on a Ramp: Understanding Fcosθ - mgsinθ = ma

  • Thread starter Thread starter frownifdown
  • Start date Start date
  • Tags Tags
    Forces Ramp
Click For Summary
SUMMARY

The discussion focuses on the equation Fcosθ - mgsinθ = ma, which represents the forces acting on an object on a ramp in physics. The x-component of the forces is defined along the ramp, with positive direction moving up the ramp. Participants emphasize the importance of understanding the derivation of these components through free body diagrams, which clarify how the forces are resolved into their respective components. This foundational knowledge is crucial for solving problems related to forces on inclined planes.

PREREQUISITES
  • Understanding of basic physics concepts such as forces and motion
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Ability to interpret and create free body diagrams
  • Knowledge of Newton's second law of motion (F = ma)
NEXT STEPS
  • Study the derivation of forces on inclined planes using free body diagrams
  • Learn about the application of trigonometric functions in physics problems
  • Explore Newton's laws of motion in greater detail
  • Practice solving problems involving forces on ramps and inclined surfaces
USEFUL FOR

Students preparing for physics exams, educators teaching mechanics, and anyone seeking to deepen their understanding of forces on inclined planes.

frownifdown
Messages
72
Reaction score
0
Here is the link to the problem http://imgur.com/9Vnc5Er

Alright, so I already know how to solve this problem, I am just struggling with the why. This is for my physics test, and he posts the answers to these in class assignments so we can review them.

Looking at step 2, we are supposed to use the equation Fcosθ - mgsinθ = ma(in the x direction). Why is this? Is this because those are the x components of the forces? I don't understand where he is getting those from. If you could shed some light on it that would be awesome. Thanks
 
Physics news on Phys.org
The x-direction used is clearly along the ramp with positive direction going up along the ramp, I hope you are not confused about how the x-y coordinate system is defined in the given solution. You might want to draw a free body diagram to clear things up. When you take component of forces along the ramp, that is what you'll arrive at.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
3K
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
20K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
5K
Replies
13
Views
5K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K