Sum of Forces in x- and y-directions

  • Thread starter jjli
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  • #1
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Homework Statement


Yusef pushes a chair of mass m = 55.0 kg across a carpeted floor with a force F of magnitude F(push) = 144 N directed at theta = 35.0 degrees below the horizontal. The magnitude of the frictional force between the carpet and the floor is F(fr) = 106 N. What is the magnitude of the acceleration a of the chair? What is the magnitude of the normal force F_N acting on the chair?



Homework Equations


\Sigma F_x = F_{\rm p}\cos\theta - F_{\rm fr} = ma_x
\Sigma F_y = F_{\rm N} - F_{\rm G}- F_{\rm p}\sin\theta = ma_y


The Attempt at a Solution



To figure out the sum of the forces in the x-direction, I plugged in known values, and found that my acceleration in the x-direction was .217416 m/s^2. I thought I used the correct method to figure out the sum of forces in the y-direction (I assumed the force of gravity and the normal force were equivalent) and I found out that the acceleration in the y-direction was -1.50173 m/s^2. However, my magnitude of acceleration, 1.517 m/s^2, was not the correct answer. The question asked for both the magnitude of acceleration and the normal force (which I figured to be 55(g): 539), so perhaps my normal force was wrong?

Help!
 

Answers and Replies

  • #2
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Hi

The first part is right. For the second part we know that the normal force (R) is equal to m*g when no other forces are acting on the chair. In this case there is an external force acting on the chair, a component of this force is acting in the upwards y direction. Yet there is no acceleration in the y direction so we know that the forces must be balanced.

So in this case:

R + (144)sin30 = m*g

You can solve for R from there.
 

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