Why Does Pulling at an Angle Affect the Normal Force on a Box?

Click For Summary
SUMMARY

The discussion focuses on calculating the normal force acting on a 30 kg box being pulled at an angle of 15 degrees with a force of 120 N on a frictionless surface. The gravitational force (Fg) is calculated as 294 N (30 kg * 9.8 m/s²). The normal force is not equal to the gravitational force due to the vertical component of the applied force, which must be subtracted from the gravitational force to determine the normal force. If the vertical component of the pulling force exceeds the gravitational force, the box will lift off the ground.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of vector components in physics
  • Familiarity with gravitational force calculations (Fg = mg)
  • Basic principles of forces on frictionless surfaces
NEXT STEPS
  • Learn how to resolve forces into horizontal and vertical components
  • Study the concept of normal force in different scenarios
  • Explore the effects of angle on force components in physics
  • Investigate conditions for an object to lift off the ground under applied forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of force analysis in real-world applications.

keemosabi
Messages
109
Reaction score
0

Homework Statement


A 30 kg box is on a frictionless floor. The box is being pulled with a force of 120 N at an angle of 15 degrees. Find the normal force. Why is the normal force not equal to simply the gravitational force. What would make the person pull the box vertically into the air?


Homework Equations


Fg = mg


The Attempt at a Solution


I solved for the horizontal component of the applied force, added it to the normal force, and set it equal to 0 plus the gravitational force (30 * 9.8 = 294 N). I think I have done the first part correctly, but I am not sure how to answer the second two parts.
 
Physics news on Phys.org
Okay, I'm a little confused with what you've done. Firstly, you solved for the horizontal component of the applied force? Why is this? You should have solved for the vertical component.

You then said that you added it to the normal force, however as defined by the question; the normal force is what we're solving for. So how exactly did you manage that?

You should solve for the vertical component of the applied force and subtract it from the magnitude of mg. This should give you the normal force.

You can see that it's not simply equal to mg because there are more forces in the y plane than simply mg.

I don't completely understand the wording of the third question, if it's asking for a numerical answer or not, but you can see that the box would leave the ground if the vertical component of applied force exceeds that of gravity.
 

Similar threads

Replies
21
Views
3K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K