Calculating Normal Force on a Box with Pulley System | Force Problem Explained

  • Thread starter Thread starter rachellehowe
  • Start date Start date
  • Tags Tags
    Force
Click For Summary
SUMMARY

The normal force exerted on a 14.8 kg box on a table, with a rope applying 3.5 Newtons of upward force, is calculated using the equation Fn = mg - T, where T is the tension in the rope. The weight of the box is 145.04 Newtons (14.8 kg * 9.8 m/s²), leading to a normal force of 141.54 Newtons (145.04 N - 3.5 N). If an 80 kg person sits on the box without any rope tension, the normal force becomes 1584.8 Newtons (80 kg * 9.8 m/s² + 145.04 N). This analysis confirms the equilibrium of forces acting on the box.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of force calculations
  • Familiarity with the concepts of weight and normal force
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the principles of static equilibrium in physics
  • Learn about tension in ropes and its applications in pulley systems
  • Explore advanced force analysis techniques in mechanics
  • Investigate the effects of additional loads on normal force calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators teaching force analysis and equilibrium concepts.

rachellehowe
Messages
2
Reaction score
0

Homework Statement


A 14.8kg box is placed on a table, and a rope, rigged to a pulley, is attached to the top of the box. Using the rope, 3.5 Newtons of force, in the upward direction (ie the rope is pulling the box up), is applied to the box.
a)what would be the normal force exerted on the box by the table?

b) if and 80 kg person sat on the box, and there was no longer a force being exerpted by the rope, what would the new normal force on the box be?


Homework Equations





The Attempt at a Solution



I tried for part a). but I'm not sure if I did it right.
F(normal)= mass*gravity
so i just multiplied 14.8kg by 9.8 m/s/s and got 145.04 Newtons then subtracted the amount from the rope and got 145.04-3.5=141.54Newtons but i don't think its right.. i need help!
 
Physics news on Phys.org
rachellehowe said:
I tried for part a). but I'm not sure if I did it right.
F(normal)= mass*gravity
so i just multiplied 14.8kg by 9.8 m/s/s and got 145.04 Newtons then subtracted the amount from the rope and got 145.04-3.5=141.54Newtons but i don't think its right.. i need help!
Sounds good to me (assuming I understand the setup).

Here's a better way to think of this problem. Account for all the forces on the box:
(1) the weight, mg, acting down
(2) the rope tension, 3.5 N, acting up
(3) the normal force from the table, Fn, acting up

These forces must add to zero, since the box is in equilibrium:
Fn + 3.5 - mg = 0

So Fn = mg - 3.5
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
30
Views
8K
Replies
46
Views
7K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K