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Five Force Problems please help

  1. Oct 1, 2009 #1
    I know it seems like five is a lot of problems to need help with but these are out of a set of 57 homework problems. I've gotten all of the others but for some reason I can't nail down these last five. Any direction you can give me on any of these problems would be appreciated.

    1. The problem statement, all variables and given/known data
    Joe pushes down the length of the handle of a 10.8kg lawn spreader. The handle makes an angle of 41.5 degrees with the horizontal. Joe wishes to accelerate the spreader from rest to 1.33 m/s in 1.5 seconds. What force must Joe apply to the handle?

    mass = 10.8 kg theta = 41.5 Vo = 0 Vf = 1.33 m/s t= 1.5s

    2. Relevant equations

    a= vf-vi / tf - ti
    F= ma

    3. The attempt at a solution

    So first I solved for the acceleration: 1.33m/s / 1.5s = 0.88667 m/s2
    I then plugged that into formula for force. (10.8kg)(0.88667m/s2) = 9.576N
    Then I took the x component of the force at the angle: 9.576cos(41.5) = 7.1719N however this is not the correct answer.

    1. The problem statement, all variables and given/known data
    A 3.17 kg ball is dropped from the roof of a building 177.8 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 13.8 N on the ball. How long does it take to hit the ground? The acceleration of gravity is 9.81 m/s2 .

    mass = 3.17kg xy = 177.8m Fx = 13.8N g = 9.81 m/s2

    2. Relevant equations
    R = sqrt[(Rx)^2 + (Ry)^2]
    F=ma
    t2 = xf-xi / 2a

    3. The attempt at a solution
    First I discovered the normal gravitational force: (3.17kg)(9.81m/s2) = 31.098N
    Then resolved the two forces in order to find the total magnitude of the force: sqrt[13.8N^2 + 31.098N^2] = 34.022N
    Then I solved for the acceleration: 34.022N / 3.17kg = 10.7327m/s2
    And finally I can find the time: sqrt[177.8/(2)(10.7327m/s2) = 2.878s which is wrong

    1. The problem statement, all variables and given/known data
    The horizontal surface on which the block of mass 4.9 kg slides is frictionless. The force
    of 54 N acts on the block in a horizontal direction. The force of 108 N acts on the block at an angle of 60 degrees as shown (south easternly). The acceleration of gravity is 9.8 m/s2. What is the magnitude of the resulting acceleration of the block?

    mass = 4.9kg Fx = 54N F=108N theta = 60 g=9.8m/s2

    2. Relevant equations
    R = sqrt[(Rx)^2 + (Ry)^2]
    F=ma

    3. The attempt at a solution
    First I broke down the force into x and y components 108cos(60)= 54N 108sin(60)=93.53N
    Then I added the forces on the x component which oppose each other so 54N - 54N = 0
    Then I solved for the total magnitude of the force sqrt[0^2 + 93.53N^2] = 93.53N
    Then I solved for acceleration 93.53N / 4.9kg = 19.0878 m/s2 also wrong


    1. The problem statement, all variables and given/known data
    A 26 kg block slides down a frictionless slope which is at angle θ = 30◦. Starting from rest,
    the time to slide down is t = 1.46 s. The acceleration of gravity is 9.8 m/s2 . What total distance did the block slide?

    mass = 26kg theta = 30 degrees Vo=0 t=1.46s g=9.8 m/s2

    2. Relevant equations
    F=mg
    -mgcostheta= ma
    x = 1/2at^2

    3. The attempt at a solution
    So first I solved for the force: (26kg)(9.8m/s2) = 2548N
    I then broke this down into components: Fx = 2548cos30=2206.6N Fy = 2548sin30=1274N
    Then I solved for acceleration: I used the y component of the force because all of the example problems of inclines I have say to do this but it just feels wrong to me 1274N / 26kg = 49 m/s2
    And now I can find the distance: 1/2(49m/s2)(1.46^2)=52.22 which I know is entirely too large.

    1. The problem statement, all variables and given/known data
    Two blocks with masses of 45.7 kg and 21.5 kg are stacked on a table with the heavier
    block on top. The coefficient of static friction is 0.650 between the two blocks and 0.270
    between the bottom block and the table. A horizontal force is slowly applied to the top
    block until one of the blocks moves. c) What minimum value for the coefficient
    of static friction between the masses and the table would cause the slippage to first happen
    between the blocks?

    mass1 = 45.7 kg mass2= 21.5kg mus(blocks)= 0.650 mus(table)= 0.270

    from part a and b I have already solved for the friction force between the blocks and between the table and I know they are correct. Between the two blocks the friction force= 291.406N. Between the table and the bottom block the friction force= 177.81N

    2. Relevant equations
    I think this a theory question but I'm not sure

    3. The attempt at a solution
    Because the coefficient of static friction between the table and the bottom block is slower I thought that was the answer. At that coefficient the bottom block will begin to slide given a lower amount of force but this is incorrect and I'm at a loss as to how to proceed.

    I realize that several of these problems seem like they need to use a similar approach and obviously it's not something I have grasped yet. I'm a visual learner and haven't been able to see too many examples of these, if you could please break it down for me it would be so helpful. Thanks for the guidance
     
  2. jcsd
  3. Oct 1, 2009 #2
    the spreader accelerates in a horizontal direction. What you computed here is the horizontal
    component of the force that joe pushes with.
     
  4. Oct 1, 2009 #3
    A horizontal force doesn't influence the falling time at all. If you have F = ma and F and are are vectors, then F and a must have the same direction, and both their horizontal and vertical components must be the same.

    this means that [tex] F_{gravitation} = m * a_{vertical} [/tex] and the horizontal component of F or a doesn't matter at all.
     
  5. Oct 1, 2009 #4
    Problem 3 I can't do without a graph.

    With problem 4 you made an error in your computation here: (26kg)(9.8m/s2) = 2548N.

    with all these slope problems you have to split all forces in components parallel and perpendicular to the slope.
     
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