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Refrigerator Question

  1. Nov 22, 2005 #1
    (1) A refrigerator with mass m is pushed up a ramp at constant speed by a man applying a force F . The ramp is at an angle [tex] \theta [/tex] above the horizontal. Neglect friction for the refrigerator. If the force F is horizontal , calculate its magnitude in terms of m and [tex] \theta [/tex].

    Ok so I drew a free body diagram. The forces acting on it are the applied force F , the weight of the crate [tex] w = mg [/tex], and the normal force n . I set up a coordinate system in which the side parallel to the ramp is the x-axis, and the side perpendicular to the ramp is the y-axis. So I decomposed the weight vector into its following components: [tex] F_{x} = mg\sin\theta [/tex], and [tex] F_{y} = mg\cos\theta [/tex]. I know the answer is [tex] mg\tan\theta [/tex]. How would you get this knowing the components of the weight vector?

    Thanks
     
  2. jcsd
  3. Nov 22, 2005 #2

    mezarashi

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    Homework Helper

    You have a good start. So the forces of interest are those along the parallel direction of the inclined ramp. By resolving the gravitational force mg, we get along the ramp direction [tex] F_g_x = mgsin\theta [/tex].

    Now, to resolve the horizontal force F being applied to the refrigerator. If we denote a force [tex]F'[/tex] to represent the component of this force parallel to the ramp's incline, then we can see the relationship: [tex]Fcos\theta = F' [/tex]. Noticeably then, for no acceleration along the ramp, then [tex] F' = F_g_x [/tex]. Solved?
     
    Last edited: Nov 22, 2005
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