Easy problem dealing with tension and mass of an object.

  • Thread starter Thread starter epicwinning97
  • Start date Start date
  • Tags Tags
    Mass Tension
AI Thread Summary
The problem involves calculating the tension force required to hold a 20kg box at rest on a frictionless ramp. The gravitational force acting on the box is determined to be 196N. The user calculated a tension force of 226N using trigonometric functions, but further clarification indicates that the correct tension force at a 30-degree angle should be 98N. Additionally, the normal force must be calculated as it acts perpendicular to the surface, requiring an understanding of the components of gravitational force. The discussion emphasizes the importance of correctly relating the forces involved to solve for tension and normal force accurately.
epicwinning97
Messages
1
Reaction score
0

Homework Statement


The box on the frictionless ramp is held at rest by the tension force. The mass of the box is 20kg. What is the value of the tension force?

Homework Equations


sine=opp/hyp cosine=adj/hyp F=ma

The Attempt at a Solution


Did 20*9.8 to get the force of gravity on the object (196N). Then I took the cosine of 30 and got .866. I then divided 196 by this and got 226N for the hypotenuse. Is this correct? Thanks!
DIAGRAM:http://img838.imageshack.us/img838/6927/diagramu.png

EDIT: So, the Newtons needed to hold it when the slope is 0 would be 196. I also found the Newtons for it at 89 degrees, which was about 11,000. So, according to my math, the tension force is 226N. Now I just need to find the normal force. Any ideas?
 
Last edited by a moderator:
Physics news on Phys.org


epicwinning97 said:

Homework Statement


The box on the frictionless ramp is held at rest by the tension force. The mass of the box is 20kg. What is the value of the tension force?

Homework Equations


sine=opp/hyp cosine=adj/hyp F=ma

The Attempt at a Solution


Did 20*9.8 to get the force of gravity on the object (196N). Then I took the cosine of 30 and got .866. I then divided 196 by this and got 226N for the hypotenuse. Is this correct? Thanks!
DIAGRAM:http://img838.imageshack.us/img838/6927/diagramu.png

Each of those calculations is correct, but now you have to relate them to the question.

EDIT: if the angle of the slope was 0o what value would you get from your calculations, and what tension would be needed to keep the mass in place?
 
Last edited by a moderator:
Mmmm... I did the same problem and got 98 N for tension. Remember that the direction of normal force is perpendicular to surface, and it isn't antiparallel to the force of gravity. You have to find the components of the force of gravity, too.
Add all components and set it equal to zero, since it is at rest.
 
The normal force is equal to a component of the force of gravity. (Try to find out which one:))
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top