How do you find the normal force of an object being pushed?

AI Thread Summary
To find the normal force of a wagon being pushed at an angle, consider the forces acting on it: gravity, friction, the applied force, and the normal force. The equation for vertical forces is set up as Fy = mgcos(46) - N, where N is the normal force. It's important to recognize that the applied force has both vertical and horizontal components due to the angle. The sum of the vertical forces must equal zero for equilibrium in the vertical direction. Understanding these components will allow for the calculation of both the normal force and the acceleration of the wagon.
Maya KDeoimmich
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Homework Statement


A 42 kg wagon is being pushed with a metal rod that makes an angle of 46 with the horizontal. The applied force is 449 N and the coefficient of kinetic friction between the wagon and the ground is 0.21.
I'm not sure what equation to use to find the normal force and then the acceleration. Thanks!

Homework Equations


Fy = mgcos46-N

The Attempt at a Solution

 
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Maya KDeoimmich said:
I'm not sure what equation to use to find the normal force and then the acceleration.
You'll use Newton's laws, of course.

First step: What forces act on the wagon? (I count three forces.)
 
Ok thanks but can you please walk me through the steps of how to find the normal force and the acceleration
 
First answer my question. What forces act on the wagon?
 
Gravity, friction, and the applied force
 
Maya KDeoimmich said:
Gravity, friction, and the applied force
Oops, one more force acts. (I meant to say four, not three.) But those three are good. What's missing?
 
normal force? not sure
 
Maya KDeoimmich said:
normal force? not sure
Exactly.

Now you'll set up two equations: One for vertical forces (vertical components) and another for horizontal forces.

Hint: What must the sum of the vertical forces equal?
 
Maya KDeoimmich said:
an angle of 46 with the horizontal
Which way? I.e. is it tending to push down on the wagon or up on it?
Maya KDeoimmich said:
Fy = mgcos46-N
How do you get that? Isn't the mg vertical?
Maya KDeoimmich said:
the coefficient of kinetic friction between the wagon and the ground is 0.21
Ouch! Unless the wheels are locked they mean rolling resistance, not friction, but don't worry about that for now.
 
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