Yazomi
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I can’t seem to get to a=4? I got 7.21=a instead. Can someone explain why/how to get to 4=a plsss
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mqb2766
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The resultant force is nonzero in the y direction so you can't balance the forces in a triangle.
Resolve the forces in x and y directions. What must happen in the x direction?
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Yazomi
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(Original post by mqb2766)
The resultant force is nonzero in the y direction so you can't balance the forces in a triangle.
Resolve the forces in x and y directions. What must happen in the x direction?
Be equal to 6n?
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mqb2766
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(Original post by Yazomi)
Be equal to 6n?
Sort of. The net force in the x direction should be 0.
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Yazomi
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(Original post by mqb2766)
Sort of. The net force in the x direction should be 0.
But I’m confused as to why that should be 0 if it’s not in equilibrium and that the y direction shouldn’t be equal?
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mqb2766
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(Original post by Yazomi)
But I’m confused as to why that should be 0 if it’s not in equilibrium and that the y direction shouldn’t be equal?
Forces in orthogonal directions (90 degrees to each other) are independent.
So it can be in equilibrium in the x direction, but have a non-stop resultant force (not in equilibrium) in the y direction.
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Yazomi
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(Original post by mqb2766)
Forces in orthogonal directions (90 degrees to each other) are independent.
So it can be in equilibrium in the x direction, but have a non-stop resultant force (not in equilibrium) in the y direction.
So to have a resultant force, the equilibrium isn’t equal? And if x is (for instance) is not in equilibrium do I automatically assume that y would be in equilibrium?
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mqb2766
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(Original post by Yazomi)
So to have a resultant force, the equilibrium isn’t equal? And if x is (for instance) is not in equilibrium do I automatically assume that y would be in equilibrium?
The question tells you that the resultant force is in the y direction, so
* The resultant force in the x direction is zero.
* The resultant force in the y direction is non-zero
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Yazomi
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(Original post by mqb2766)
The question tells you that the resultant force is in the y direction, so
* The resultant force in the x direction is zero.
* The resultant force in the y direction is non-zero
Ahhhhhh trying to get me head around mechanics, I got you now tho thanks!! Finally made sense
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mqb2766
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(Original post by Yazomi)
Ahhhhhh trying to get me head around mechanics, I got you now tho thanks!! Finally made sense
Spot the independent ball motion towards the end
https://www.youtube.com/watch?v=fm5C4FvI6Po&app=desktop
The vertical (ball) motion does not affect the horizontal (ball) speed.
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Yazomi
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(Original post by mqb2766)
Spot the independent ball motion towards the end
https://www.youtube.com/watch?v=fm5C4FvI6Po&app=desktop
The vertical (ball) motion does not affect the horizontal (ball) speed.
It would be the horizontal direction wouldn’t it,

If it’s large scale enough friction would affect it wouldn’t it
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mqb2766
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(Original post by Yazomi)
It would be the horizontal direction wouldn’t it,

If it’s large scale enough friction would affect it wouldn’t it
Sure, it's a simple example where the vertical force (shooting the ball upwards) does not affect the horizontal motion (constant velocity). As with most examples, the real world can get messy (friction, ...).
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