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    Lets say i have a box with a weight of 20N. If i want to push the box vertically upwards i would have to apply 20N or more force upwards. If i want to move it upwards at an angle, i.e. apply an upwards force at an angle of say 60degrees from the ground, the total force i would need to apply to the box would be 20/sin(60), is that right?
    This means a force of 23N is required to move it upwards at an angle of 60degrees. I would've thought however that moving it upwards at an angle would require less force? So is it my thinking that is wrong or my calculation?
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    Hey Whispering eye (PS fantastic name ;D),

    I can understand what you mean in that one could think that it should be less, but you calculation was correct, you did draw yourself a diagram? diagrams always help to properly visualise a system.

    Have a think about this though, if you applying a force at an angle such that the vertical component of the applied force, in this case 'pushing the box', is greater than its weight it will move (remember if the box is currently at rest you would have to apply a force greater than but not equal to 20N ), or more precisly accelerate upwards, but you will also be causing the box to move horizontaly too. Thinking about that, does it make sense that you would require more than 20N to simply move the box vertically, but you would require less to both move the box vertically and horizontally.

    Hope thats helps Whispering
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    Note that if you are thinking "pushing a box up a slope", then the slope is providing most of the vertical force.
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    (Original post by Galadirith)
    Hey Whispering eye (PS fantastic name ;D),

    I can understand what you mean in that one could think that it should be less, but you calculation was correct, you did draw yourself a diagram? diagrams always help to properly visualise a system.

    Have a think about this though, if you applying a force at an angle such that the vertical component of the applied force, in this case 'pushing the box', is greater than its weight it will move (remember if the box is currently at rest you would have to apply a force greater than but not equal to 20N ), or more precisly accelerate upwards, but you will also be causing the box to move horizontaly too. Thinking about that, does it make sense that you would require more than 20N to simply move the box vertically, but you would require less to both move the box vertically and horizontally.

    Hope thats helps Whispering
    Ye i see what you mean, it does make sense. What if i push it upwards at angle of just 1degree, 20/sin(1) = 1146N! Why would the force be so much in this case? Its hardly moving upwards against gravity now yet so much force is needed just for the horizontal movement?
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    (Original post by DFranklin)
    Note that if you are thinking "pushing a box up a slope", then the slope is providing most of the vertical force.
    Nah no slope involved
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    (Original post by Whispering Eye)
    Ye i see what you mean, it does make sense. What if i push it upwards at angle of just 1degree, 20/sin(1) = 1146N! Why would the force be so much in this case? Its hardly moving upwards against gravity now yet so much force is needed just for the horizontal movement?
    You need 20 N vertical just to stop it falling due to gravity. Since only a tiny part of the force you're applying is in the vertical direction, the total force you apply will have to be large.
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    (Original post by DFranklin)
    You need 20 N vertical just to stop it falling due to gravity. Since only a tiny part of the force you're applying is in the vertical direction, the total force you apply will have to be large.
    If i only apply a horizontal force, so the angle is 0degrees, then the force required is f=ma right? So then why if i only increase the angle by 1degree does the force shoot right up? Sorry if these are stupid questions, i just really want to get my head around it!
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    If you apply a horizontal force however large, the box is going to fall.
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    (Original post by Whispering Eye)
    If i only apply a horizontal force, so the angle is 0degrees, then the force required is f=ma right? So then why if i only increase the angle by 1degree does the force shoot right up? Sorry if these are stupid questions, i just really want to get my head around it!
    That's not right.

    Just to clarify: when you say 'push upwards', do you mean you're lifting the box off a surface? Or is it suspended from a rope or something? Or in an area with negligible gravitational effects? Like DFranklin says, if there's no vertical supporting force at all (like if the force were acting completely horizontally), the mass of the block will not be suspended and it will fall due to gravity. :yep:

    Draw a diagram; consider the vertical and horizontal components of the force. You can express these components in terms of the applied force and the angle at which said force is being applied, by using trigonometry. :yes:
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    The box is connected to a beam, which is connected to a hydraulic. The hydraulic applies a push force to the beam. So when the box is at a horizontal position, the beam too is horizontal but it is attached to the box, so the box wont fall down. I am trying to find at what condition the hydraulic will apply maximum force. When everything is horizontal, the force it must apply to move the beam and box is f=ma. And when applying a force at an angle, it is f = total mass of beam and box / sin(angle). Is this right so far? If so, then at what condition would max force be?.. When the angle tends to 0? i.e. when its at horizontal? But moving it at horizontal is not going against gravity, so how can this be? Ahhh... im so confused!
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    I think you need to post a diagram. I have no idea what your hydraulic is actually doing here.
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    If you assume that the angle is fixed, that everything is infinitely rigid, and there's no friction in the hydraulic, then the force is simply mg sin a.

    (Essentially because the infinitely rigid claim is effectively the same as putting it all on a slope).
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    The hydraulic can move from 0 to 90degrees. I am trying to find the angle at which the maximum force is needed by the hydraulic to push the beam.
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    Yes, but you're not going to change the angle during the push, are you?
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    (Original post by DFranklin)
    If you assume that the angle is fixed, that everything is infinitely rigid, and there's no friction in the hydraulic, then the force is simply mg sin a.

    (Essentially because the infinitely rigid claim is effectively the same as putting it all on a slope).
    Why would that be the force? wouldnt it be F = mg/sin(a)?
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    No.
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    Lol, why not?
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    because the infinitely rigid claim is effectively the same as putting it all on a slope
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    why is the force not mg/sina ?

    sina = mg/F no?
 
 
 
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