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Block pushed up wall

  1. Oct 14, 2004 #1
    Hey everyone, my first post here, hope it's ok that it's a question...
    I've got this physics problem that is absolutely stumping me, is abnormal from all the other forces problems I've done before, and uk is throwing me for a loop too...here it is:

    6. [SFHS99 5.P.60.] A 5.5 kg block is pushed 3.1 m at a constant speed up a vertical wall by a constant force applied at an angle of 30.0° with the horizontal, as shown in Figure 5-23.
    Figure 5-23
    Assume that the coefficient of kinetic friction between the block and the wall is 0.30.

    (a) Determine the work done by the force on the block
    ___________________ J
    (b) Determine the work done by gravity on the block.
    ___________________ J
    (c) Determine the normal force between the block and the wall.
    ___________________ N

    Any help or hints you guys can give me would be greatly appreciated...this one has me stumped.
     
  2. jcsd
  3. Oct 14, 2004 #2

    Pyrrhus

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    You know there will be a normal force, find out which will it be equal to, remember the block isn't moving the x-axis so the Fx sum is zero.
    Also Remember Newton's 1st Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{v} = constant[/tex]

    Think in components of the applied force.
     
    Last edited: Oct 14, 2004
  4. Oct 14, 2004 #3
    Ok so I tried taking the gravitational force, 54.96N, and finding the horizontal component of that. I drew it as a triangle with 54.96N as the y leg, then did 53.96/tan 30 to find the x leg of the triangle. This gave me 93.46N, which my online homework is saying is wrong! Still confused =\
     
  5. Oct 14, 2004 #4

    Pyrrhus

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    Applying Newton's 1st Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{v} = constant[/tex]

    on x-axis

    [tex] Fcos\theta - N = 0 [/tex]

    [tex] Fcos\theta = N [/tex]

    On y-axis

    [tex] Fsin\theta - mg - F_{f} = 0 [/tex]

    [tex] F_{f} = \mu N [/tex]

    [tex] Fsin\theta - mg -\mu Fcos\theta = 0 [/tex]

    Gravity will have a negative work because it forms 180 degrees with the displacement vector, and [itex] Fsin\theta [/itex] will do the work pushing the block up.

    Also remember Work definition

    [tex] W = \vec{F} \cdot \vec{r} [/tex]

    where W is equal to

    [tex] W = |\vec{F}||\vec{r}|cos\theta [/tex]

    where [itex] \theta [/itex] is the angle between the radius vector and the force.
     
    Last edited: Oct 14, 2004
  6. Oct 14, 2004 #5
    ok I've gotten the Fapp and the Fn, now I'm just having trouble with the Ff. I used the equation Ff = uN, plugging in my normal force of 194.5 and u of .3, then multiplied that by 3.1 to get work, and entered the negative of that, but it's still telling me its incorrect. Don't understand what I'm doing wrong!
     
  7. Oct 14, 2004 #6
    Didn't see your new post and edit, might be able to get it with that.
     
  8. Oct 14, 2004 #7

    Pyrrhus

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    Read the questions again, it does not ask for work done by friction... I also decided to delete my energetic solution, because it wasn't needed.
     
  9. Oct 14, 2004 #8
    Whoops it was work done by gravity...easy just mgd. doh! Thanks for the help i got them all now.
     
  10. Oct 14, 2004 #9

    Pyrrhus

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    It was a pleasure to be of help, and welcome to PF!
     
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