When normal force isn't normal l

AI Thread Summary
The discussion focuses on understanding how to calculate the normal force acting on a box being pulled at an angle by a rope. The normal force is not simply the weight of the box but is affected by the upward component of the tension in the rope. Specifically, the normal force can be expressed as the weight of the box minus the vertical component of the tension, which is Tsin(theta). Participants clarify that the normal force changes based on the angle of the rope and the tension applied. This highlights the importance of considering all forces acting on the box to accurately determine 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 \theta with the horizontal and has tension T, then the "upward force" is Tsin(\theta) and the horizontal force is Tcos(\theta). The "normal force" (normal to the surface) is the weight of the box, mg, minus that upward force: mg- Tsin(\theta).
 
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