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Sum of Forces in x- and y-directions

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    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?



    2. Relevant 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


    3. 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!
     
  2. jcsd
  3. Sep 19, 2009 #2
    Note that if the chair is not moving anywhere in the y-direction, then it does not have a net acceleration in the y-direction...unless the question is asking for the acceleration done by a specific force.

    The applied force is 35 deg below the horizontal, so that would mean you have part of the force in the vertically downward direction, and part in the horizontal direction. The downward component, in addition to the weight of the chair, push against the floor; therefore, the normal force is the sum of the downward component and weight and is in the vertically upward direction - the normal force follows Newton's third law: If a force A is exerted on the floor, then the floor exerts a force B that is equal and opposite to force A.
     
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