How Do You Calculate Forces on Inclined Planes?

  • Thread starter Thread starter jarny
  • Start date Start date
  • Tags Tags
    Forces
Click For Summary

Homework Help Overview

The discussion revolves around calculating forces acting on blocks on inclined planes, specifically involving tension in cords and normal forces. The original poster presents two distinct problems involving different masses and angles on ramps.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate tension and normal forces using gravitational components and has shared specific numerical attempts. Some participants question the accuracy of the calculations and the need for diagrams to clarify the problems.

Discussion Status

Participants are exploring various components of forces acting on the blocks, with some guidance offered regarding the relationship between tension and gravitational forces. There is an ongoing exchange about the correct approach to resolving the forces involved.

Contextual Notes

The original poster mentions revisions to their answers, indicating uncertainty about the correctness of their calculations. There is also a reference to the need for diagrams, which may affect the clarity of the problem setup.

jarny
Messages
15
Reaction score
0

Homework Statement



I have two questions:

1. Let the mass of the block be 8.2 kg and the angle θ be 37°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the block's acceleration.

2.In the figure, a crate of mass m = 104 kg is pushed at a constant speed up a frictionless ramp (θ = 29°) by a horizontal force f-> . The positive direction of an x-axis is up the ramp, and the positive direction of a y-axis is perpendicular to the ramp. (a) What is the magnitude of f-> ? (b) What is the magnitude of the normal force on the crate?



Homework Equations



F_net=ma
Maybe T-m*g*sin(theta)=ma


The Attempt at a Solution



Ok, for 1.a. I used 8.2kg*9.8m/s^2*cos(29)=48.3619N to find the tension on the cord. For b I used 8.2*9.8*sin(29)=64.1783N. For 3. I am having trouble because my T and mgsintheta are the same.

For 2a. I used 104kg*9.8m/s^2tan(29) to get 564.92N and for B is got 1019N*cos(29) but the computer says this is wrong. The only one I got right was 2.a, 1.a and 1.b were revised so they might be right.



Can anyone help? Thank you very much
 
Physics news on Phys.org
Welcome to PF.

What is the diagram of the problems you are solving.
 
1)The force component of gravity pointing down the ramp is sinθ.

Hence the Tension to off set that in the direction up the ramp is m*g*sinθ = m*a where a is the answer to your part c.

The Cosθ component of g determines the force normal to the incline.

2) For part b you have two components of force to account for.

From the F you have F*sinθ
From mass and gravity you have m*g*Cosθ
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
46
Views
7K