Magnitude of Force within incline?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a block on a frictionless incline, where participants are trying to determine the magnitude of a horizontal force and the normal force acting on the block. The incline is set at an angle of 34 degrees, and the mass of the block is given as 5.5 kg.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the block, including the use of sine and cosine functions to resolve the forces along the incline and perpendicular to it. There are attempts to clarify the correct angles to use in calculations, with some confusion about the application of trigonometric functions. Questions arise regarding the accuracy of previous calculations and the significance of rounding in their answers.

Discussion Status

There is ongoing exploration of different approaches to solving the problem, with participants sharing their calculations and questioning the correctness of their results. Some participants suggest that the horizontal force does not equal the force parallel to the incline, indicating a need for further clarification. Multiple interpretations of the problem setup are being discussed, and guidance has been offered regarding the use of angles in calculations.

Contextual Notes

Participants note the potential impact of rounding on their answers and express uncertainty about the correctness of their calculations. There is mention of an online platform that provides immediate feedback on answers, which may influence their approach to solving the problem.

  • #61
anglum said:
Fn -53.9cos(34) -20.3 = 0
Fn = 53.9cos(34) + 20.3
Fn = 44.685 + 20.3
Fn = 64.9851N

yes. that's correct.
 
Physics news on Phys.org
  • #62
Fn- 54cos(34) - 20.3 = 0
Fn - 44.7 - 20.3= 0 <<----
Fn= 24.4

Edit: Nevermind I know what I did...
 
Last edited:
  • #63
ok that is the correct answer

so our problem was we weren't calculating all 3 perpendicular forces... we were just finding one of them and using that as Fn...

however its all 3 of them equal to 0
 
  • #64
I'm so confused... I'm going to go read over all that again haha..
 
  • #65
anglum said:
ok that is the correct answer

so our problem was we weren't calculating all 3 perpendicular forces... we were just finding one of them and using that as Fn...

however its all 3 of them equal to 0

Yes, it is important to go with the basic equations, and get the results from there... don't take shortcuts...

the most important step is \Sigma{Fy} = 0, where y is perpendicular to the plane... plug in your forces into the left side... Fn - mgcos(34) - Fsin(34) = 0...
 
  • #66
Ohhh
 
  • #67
So Fn - Fg parallel - Fg perpendicular = 0 ?
 
  • #68
thanks to all who have helped me ... tonite i finished earlier than usual... i got the other 9 problems sumwhat easily... thanks again
 
  • #69
Thanks for posting it I learned something too. Thanks learningphysics..
 

Similar threads

Calculating Net Force on a Wheel: Magnitude and Direction | 0.350 m Radius
  • · Replies 4 ·
Replies
4
Views
2K
Net Force acting on block on incline
  • · Replies 1 ·
Replies
1
Views
2K
A frictionless incline with a block sliding down it....
  • · Replies 2 ·
Replies
2
Views
3K
Physics Homework -- block pressed against a wall is sliding....
  • · Replies 4 ·
Replies
4
Views
3K
Block on an incline and frictional force
  • · Replies 9 ·
Replies
9
Views
14K
Force of a block on an incline above a horizontal
  • · Replies 10 ·
Replies
10
Views
2K
Help understanding answers: kinetic friction
  • · Replies 20 ·
Replies
20
Views
3K
What is the change in kinetic energy of the crate pulled up a rough incline?
  • · Replies 6 ·
Replies
6
Views
3K
Frictional force on an inclined plane
  • · Replies 24 ·
Replies
24
Views
5K
Static friction of a block on an incline?
  • · Replies 1 ·
Replies
1
Views
2K