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Phy 121
Your 'cq_1_17.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A 5 kg cart rests on an incline which makes an angle of 30 degrees with the horizontal.
•Sketch this situation with the incline rising as you move to the right and the cart on the incline. Include an x-y coordinate system with the origin centered on the cart, with the x axis directed up and to the right in the direction parallel to the incline.
The gravitational force on the cart acts vertically downward, and therefore has nonzero components parallel and perpendicular to the incline.
Sketch the x and y components of the force, as estimate the magnitude of each component.
What angle does the gravitational force make with the positive x axis, as measured counterclockwise from the positive x axis? Which is greater in magnitude, the x or the y component of the gravitational force?
answer/question/discussion: ->->->->->->->->->->->-> :
This is all a bit confusing to me still, but I am trying to grasp this concept.
As far as my graph goes, I have the angle that gravitational force makes with the positive x axis at 240 degrees. I assume we must take the L of 5 kg and multiply it by the sin of 60 degrees to find the y component, which is 4.3 and the x component will be the L * cos(60 degrees) = 2.5. Therefore if I did this correctly, the y component has a greater magnitude.
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Your reasoning is good, but you can't use the 60 degree angle.
Angles need to be measured in the counterclockwise direction from the positive x axis.
So you need to use the 240 degree angle, which you have correctly identified.
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•Using the definitions of the sine and cosine, find the components of the cart's weight parallel and perpendicular to the incline.
answer/question/discussion: ->->->->->->->->->->->-> :
I am not sure how to find this. If I am thinking correctly, this essentially means the x and y components, right? Therefore the y component is 4.3 and the x component is 2.5, as I found it in the previous problem.
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The y component of the cart's weight is clearly downward, in the negative y direction. It should also be clear that the x component is in the negative x direction. So your posiitve answers are numerically correct, but have the wrong sign.
Had you used the angle 240 degrees this would have been taken care of automatically, and your results would match your picture.
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•How much elastic or compressive force must the incline exert to support the cart, and what is the direction of this force?
answer/question/discussion: ->->->->->->->->->->->-> :
Would we just take the force of gravity here? If so, the force would be 5 kg * 9.8 m/s^2 = 49 Newtons, and the direction would be 240 degrees in the counterclockwise direction.
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Only the y component of the gravitational force is perpendicular to the incline, so only this component pushes "into" the incline, and only this force deforms the incline and causes it to push back.
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•If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
I would say just 9.8 m/s^2, but this may be wrong to assume.
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The component of the acceleration perpendicular to the incline does not act in the direction of motion, which is parallel to the incline.
Only the component of the force in the direction of motion has a direct effect on the acceleration.
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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
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