When normal force isn't normal l

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Homework Help Overview

The discussion revolves around understanding the normal force acting on a box being pulled across a horizontal surface by a rope at an angle. Participants are exploring the relationship between the forces involved, particularly how the normal force is affected by the tension in the rope and the weight of the box.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify how to calculate the normal force when a rope is pulling the box at an angle. Questions are raised about the components of the forces acting on the box, particularly regarding the upward component of the tension in the rope and its effect on the normal force.

Discussion Status

Some participants are questioning the assumptions made about the direction of the forces and the definition of the normal force. Guidance has been offered regarding the calculation of the normal force as the weight of the box minus the upward component of the tension in the rope, indicating a productive exploration of the topic.

Contextual Notes

There is a focus on the equilibrium of forces acting on the box, and participants are encouraged to consider different scenarios, such as varying the tension in the rope and its angle, to understand the implications for the normal force.

Gersty
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When normal force isn't "normal"l

Homework Statement


A box with a certain mass is pulled across a horizontal surface by a rope at an angle. Figure out the components, the friction, the acceleration of the box, when the tension in the rope, its angle, the mass of the box, and mu are given.


Homework Equations


F=ma, sin (theta) hypotenuse, etc...


The Attempt at a Solution


I understand the solution but I'm having trouble explaining how the normal force is found. I am used to calculating the using m(g) to calculate the force of gravity. It seems as though the normal force should just be the same as gravity plus the upward component of the rope. I get that equilibrium plays a role but am just having difficulty explaining it.
 
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Gersty said:

Homework Statement


A box with a certain mass is pulled across a horizontal surface by a rope at an angle. Figure out the components, the friction, the acceleration of the box, when the tension in the rope, its angle, the mass of the box, and mu are given.

Homework Equations


F=ma, sin (theta) hypotenuse, etc...

The Attempt at a Solution


I understand the solution but I'm having trouble explaining how the normal force is found. I am used to calculating the using m(g) to calculate the force of gravity. It seems as though the normal force should just be the same as gravity plus the upward component of the rope. I get that equilibrium plays a role but am just having difficulty explaining it.

What do you mean with the sentence "It seems as though the normal force should just be the same as gravity plus the upward component of the rope."? The rope has no "upward component". In what direction is the rope pulled?

ehild
 
Imagine a box of mass 'm' just sitting on the ground. What is the normal force exerted by the ground on the box?

Now, imagine the box has a rope tied to it. Does the normal force on the box change?

If you pull straight up on the rope so that there is a tension of 5 N and the box is still sitting on the ground, what is the normal force exerted by the ground on the box? If the tension is 10 N? When is the normal force = 0?
 
Gersty said:

Homework Statement


A box with a certain mass is pulled across a horizontal surface by a rope at an angle. Figure out the components, the friction, the acceleration of the box, when the tension in the rope, its angle, the mass of the box, and mu are given.
If the rope makes angle [itex]\theta[/itex] with the horizontal and has tension T, then the "upward force" is [itex]Tsin(\theta)[/itex] and the horizontal force is [itex]Tcos(\theta)[/itex]. The "normal force" (normal to the surface) is the weight of the box, mg, minus that upward force: [itex]mg- Tsin(\theta)[/itex].
 

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