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PHY 201
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: ->->->->->->->->->->->-> :
Not sure how to visualize this, i have drawn it out.
I have a 30 deg incline with a cart sitting on the bottom of the incline with the origin of the x and y axis going through it. Gravitational force is acting down on it, the normal force that is perpendicular to the cart and exerted by the incline, and possibly a third force acting on it for friction.
The angle that the gravitational force makes with positive x axis is 270 deg - 30 deg = 240 deg.
The y component of the gravitational force is greater.
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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: ->->->->->->->->->->->-> :
weight = mg = 5 kg * 9.8 m/s^2 = 49 N
Weight parallel = W sin 'theta = 49 N sin 240 deg = -42.4 N
Weight Perpendicular = W cos 'theta = 49 N cos 240 = -24.5 N
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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: ->->->->->->->->->->->-> :
The direction of the elastic/compressive force is perpendicular to the cart. With no external forces, this force will be equal and opposite to the perpendicular weight of the cart, therefore, it would be 24.5 N
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• If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
9.8 m/s^2
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The acceleration is important.
It isn't 9.8 m/s^2. That would be the acceleration of the cart in free fall.
What is its acceleration along the ramp?
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This is my revision. I fixed my parallel and perpendicular weight angles.
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Good, but you still need to find the acceleration. Check my notes.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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