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Ahh!! Man pushes block down an incline, with a force PARALLEL TO FLOOR Watch

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    I worked out the force parallel to the incline if he pushed it down the incline, but cos his force is now parallel to the floor, I'm stuck!!

    This is qu 5.9 from 13th edition of Uni Physics Freedman & Young:

    A man pushes a piano (180kg) so it has a constant velocity down a ramp inclined at 11.0degrees. Friction is negligible/non-existent for this question. Calculate the magnitude of the force applied by the man if he pushes:
    a) parallel to the incline:
    b) parallel to the floor:

    My answer for part a) is 336.9N

    I used W=mg=180*9.81=1765.8N, and this is the weight of the piano.
    Then, splitting the force component up using sin and cos, I saw that the force I need is in the sin11degrees direction. Force(push) = Wsin11deg. = 1765.8*sin11deg. = 336.9N

    I thought I could work out part b) using (Fcos11)^2 + (Fsin11)^2 = F^2, and rearrange to get Fcos11 = 1733.4N, but this is waaaayyyyy wrong. It looks pretty wrong - the man shouldn't need like 5 times the force if he pushes parallel to the floor.....


    Help please?

    Merci!!
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    (Original post by PhysicsGal)
    I worked out the force parallel to the incline if he pushed it down the incline, but cos his force is now parallel to the floor, I'm stuck!!

    This is qu 5.9 from 13th edition of Uni Physics Freedman & Young:

    A man pushes a piano (180kg) so it has a constant velocity down a ramp inclined at 11.0degrees. Friction is negligible/non-existent for this question. Calculate the magnitude of the force applied by the man if he pushes:
    a) parallel to the incline:
    b) parallel to the floor:

    My answer for part a) is 336.9N

    I used W=mg=180*9.81=1765.8N, and this is the weight of the piano.
    Then, splitting the force component up using sin and cos, I saw that the force I need is in the sin11degrees direction. Force(push) = Wsin11deg. = 1765.8*sin11deg. = 336.9N

    I thought I could work out part b) using (Fcos11)^2 + (Fsin11)^2 = F^2, and rearrange to get Fcos11 = 1733.4N, but this is waaaayyyyy wrong. It looks pretty wrong - the man shouldn't need like 5 times the force if he pushes parallel to the floor.....


    Help please?

    Merci!!
    I don't know if I'm misreading the question, but surely if friction is non-existent then the piano would slide down the ramp all by itself and any applied force would cause it to accelerate even faster!

    Or is the man pushing against the piano to stop it sliding too quickly?
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    (Original post by davros)
    I don't know if I'm misreading the question, but surely if friction is non-existent then the piano would slide down the ramp all by itself and any applied force would cause it to accelerate even faster!

    Or is the man pushing against the piano to stop it sliding too quickly?
    Ahhh you do make sense!! There was no diagram provided but I'll rereard the question later (my laptop is way too hot atm so I need to turn it off and let it cool down asap) and see if I read it wrong myself!
    That does make sense re frictionless!
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    (Original post by PhysicsGal)
    I worked out the force parallel to the incline if he pushed it down the incline, but cos his force is now parallel to the floor, I'm stuck!!

    This is qu 5.9 from 13th edition of Uni Physics Freedman & Young:

    A man pushes a piano (180kg) so it has a constant velocity down a ramp inclined at 11.0degrees. Friction is negligible/non-existent for this question. Calculate the magnitude of the force applied by the man if he pushes:
    a) parallel to the incline:
    b) parallel to the floor:

    My answer for part a) is 336.9N

    I used W=mg=180*9.81=1765.8N, and this is the weight of the piano.
    Then, splitting the force component up using sin and cos, I saw that the force I need is in the sin11degrees direction. Force(push) = Wsin11deg. = 1765.8*sin11deg. = 336.9N

    I thought I could work out part b) using (Fcos11)^2 + (Fsin11)^2 = F^2, and rearrange to get Fcos11 = 1733.4N, but this is waaaayyyyy wrong. It looks pretty wrong - the man shouldn't need like 5 times the force if he pushes parallel to the floor.....


    Help please?

    Merci!!
    I am truly confused as to why you used what looks like the Pythagorean theorem. Your equation is technically correct, but it gives you no new information because you do not know what F actually is for part (b); that is what you are trying to find. You need to equate F \cos 11 with the component of the weight down the incline, which is exactly what you calculated in part (a).

    Your intuition is correct, of course. It is only an incline of 11 degrees so the force required is not substantially larger.
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    (Original post by Brister)
    I am truly confused as to why you used what looks like the Pythagorean theorem. Your equation is technically correct, but it gives you no new information because you do not know what F actually is for part (b); that is what you are trying to find. You need to equate F \cos 11 with the component of the weight down the incline, which is exactly what you calculated in part (a).

    Your intuition is correct, of course. It is only an incline of 11 degrees so the force required is not substantially larger.
    Someone else said that too...now I'm not so sure. I think I used a rearrangement because I was working out the size of a different side of the triangle (the triangle that splits force components up into 2 compo.'s at right angles to each other)..Anyway..(btw why can't I create paragraphs on here anymore? o.O)....BTW why would I equate Fcos11 and weight down the incline?...anyway, so: Fcos11 = Weight down incline (which is 336.9N)......F = 336.9/cos11 = 343.2N!! Which is correct! Woo, thank you!! How did that work?!!
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    (Original post by PhysicsGal)
    Someone else said that too...now I'm not so sure. I think I used a rearrangement because I was working out the size of a different side of the triangle (the triangle that splits force components up into 2 compo.'s at right angles to each other)..Anyway..(btw why can't I create paragraphs on here anymore? o.O)....BTW why would I equate Fcos11 and weight down the incline?...anyway, so: Fcos11 = Weight down incline (which is 336.9N)......F = 336.9/cos11 = 343.2N!! Which is correct! Woo, thank you!! How did that work?!!
    There should obviously be no motion perpendicular to the plane so the normal reaction R will necessarily equal the components of the piano's weight and the force F in that direction. In order to maintain a constant velocity parallel to the incline, there must be no net force in that direction. That is why you need the component of F parallel to the incline to cancel the corresponding component of weight.
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    If the man is opposing the action of weight, then the values should come out as 336.59 N (roughly 337 N) and 342.89 N (roughly 343 N), if g=9.8 ms-2.

    However, if the man is assisting the action of weight, then without resistive forces there is no way the velocity could be constant (rather the block would be accelerating).
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    I read that as a Daily Mail headline
 
 
 
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