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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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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: ->->->->->->->->->->->-> :
5 kg * 9.8 m/s^2 = 49 N
The x axis is 30 degrees above horizontal.
The gravitational force is directed vertically downward, which puts it at 90 degrees relative to the horizontal.
The Y component has a great magnitude than the x component.
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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: ->->->->->->->->->->->-> :
The x component is magnitude * cos(theta) = 49 N * cos(240 degrees) = -24.5 N
The y component is magnitude * sin(theta) = 49 N * sin(240 degrees) = -43 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 y component of the gravitational force is -43 N
The x component of the gravitational force is -24.5 N
I’m not sure where to go from here.
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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 net force on the cart will be -24.5 N in the x direction
a = F_net / m = -24.5 N / (5 kg)= -4.9 m/s^2
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Solution
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