Pulling box at an angle

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when you pull a box at an angle is it accelerating in the y direction? I would think it is since there is now an unbalanced force in the y direction.
 

Doc Al

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when you pull a box at an angle is it accelerating in the y direction? I would think it is since there is now an unbalanced force in the y direction.
That depends on how hard you pull. It could very well be that the net force in the y direction remains zero. (Note that the normal force of the floor on the box will adjust itself as the angled force is applied.)
 

sophiecentaur

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To be sure of what you are actually asking, perhaps you could describe, in 3D?, which direction is which. Is this box just on a plane and is the force parallel to the plane?
 
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its a 2D problem and no the force is at an angle to the plane. its being pulled along the x axis.
 

sophiecentaur

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I could give the obvious answer and say that I wouldn't expect any acceleration in the y direction if there were no component of impressed force in that direction.
That presupposes that the surface is uniform, though. If there are any 'diagonal' features in the textures of the surfaces then that could produce a y component of force - and some y acceleration. 'Mating' grooves on both surfaces at 45 degrees could produce equal movement in x and y directions - that would be an extreme case..
 

Doc Al

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its a 2D problem and no the force is at an angle to the plane. its being pulled along the x axis.
If this is a typical textbook problem where a box is being pull along a horizontal surface by some rope at an angle, you can most likely assume that the box remains in contact with the surface and that there's no vertical acceleration.
 

arildno

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As Doc Al said:
The effect of having some vertical component of the pulling force is to alter the magnitude of normal force.

Clearly, it IS possible that this vertical component becomes so large as to make the box tilt, but even in the case when the box doesn't visibly get this type of vertical acceleration, the normal force has changed.
 

sophiecentaur

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I totally misread the op. Durr!

If the string is pulled so that the acceleration is enough for the vertical component to be g then the block will leave the plane. If the angle between the string and the plane is θ then the force would need to be >mg/sin(θ)
 
i am fairly new to this site, but as Doc Al said this is a typical physics problem even if the box is being pulled forward or backward at an angle it is safe to assume that the box remains on the surface or (ground) giving it no vertical acceleration.
 
I agree with Doctor AL on this question, while pulling the box on an angle you need to consider that it is still on the horizontal x plane, and therefore has no vertical acceleration. Think of it as if you were pushing or pulling a box up a hill and that might make the direction of the forces easier to visualize
 

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