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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: ->->->->->->->->->->->-> :
Gravitational force makes a 180 degree angle ( directly South ) with the x-axis, if the x-axis is pointing North ( up ) as suggested. Otherwise, it makes a 90 degree angle with x-axis, if it is positioned as normal. The x component of the gravitational force, although negative, is greater in magnitude than the y component. There is no acceleration in the horizontal direction. The gravitational force has no effect on the horizontal motion of the 5 kg cart.
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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: ->->->->->->->->->->->-> :
sin ( 30 degrees ) = 9.8 / H
H = 9.8 m/ ( sin( 30 degrees ))
H = 19.6
cos( 30 degrees ) = A/H
A = H * cos( 30 degrees )
A = 19.6 * .86660254038
A = 16.97 = 17.0
y = -17.0 ( parallel to incline if x is upward, north and southbound )
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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 positive force would have to be up or north, to combat the negative, downward force of gravity. Because the force is 49 N going in the downward direction. The force would have to be 100%, 49 N, elastic to keep the cart on the incline at equilibrium.
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• If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
The cart's acceleration will be -9.8 m/s/s ( if upward is positive ) if no other force is exerted parallel to the incline.
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Solution
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