Sum of Forces in x- and y-directions

[jjli]

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!

 

[Kalvarin]

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.

 

[tomcue]

Hello Yusef,

Thank you for providing all the necessary information for this problem. I can see that you have correctly used the equations for the sum of forces in the x- and y-directions. However, there seems to be a mistake in your calculation for the normal force.

To calculate the normal force, we need to consider all the vertical forces acting on the chair. These include the weight of the chair (mg) and the vertical component of the pushing force (Fp sinθ). Therefore, the equation for the sum of forces in the y-direction should be:

ΣFy = F_N - mg - Fp sinθ = ma_y

Using this equation and plugging in the known values, we can solve for the normal force as:

F_N = mg + Fp sinθ - ma_y = (55.0 kg)(9.8 m/s^2) + (144 N)sin(35.0°) - (55.0 kg)(-1.50173 m/s^2) = 539 N

This is the correct value for the normal force acting on the chair. Now, to find the magnitude of the acceleration, we can use the equation:

a = √(ax^2 + ay^2) = √((0.217416 m/s^2)^2 + (-1.50173 m/s^2)^2) = 1.518 m/s^2

Therefore, the magnitude of the acceleration is 1.518 m/s^2.

I hope this helps clarify any confusion you had with the problem. Keep up the good work in your studies!

Best,