PHY 121
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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: ->->->->->->->->->->->-> :
The gravitational force makes a 60 degree angle with the positive x axis, as measured counterclockwise from the positive x axis. The y component of gravitational force is greater in magnitude than the x component.
60 degrees is the nearest angle between the x axis and the gravitational force.
However that angle is with the negative x axis.
measured counterclockwise from the positive x axis the angle is 240 degrees.
• 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: ->->->->->->->->->->->-> :
Making a triangle using the x and y axis we have drawn, we can use sin(theta) = opp/hyp. And cos(theta) = adj/hyp to determine that both the horizontal and vertical components are equal.
It is possible to use right triangle trigonometry here, as long as you are very careful about the signs of the components.
Whether your expressions are appropriate depends on which angle you are calling theta, the angle of the incline, the angle with the x axis, or the counterclockwise angle with the positive x axis.
• 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: ->->->->->->->->->->->-> :
We calculate the component of gravity parallel to the hill by using m*g*sin(theta) = 5kg*9.8m/s*sin30 = .51kg m/s * sin30 = .26 N. The direction of the compressive force is perpendicular to the incline, and it is equal and opposite the weight of the cart, making it -.26N.
The compressive force is a response to the gravitational component perpendicular to the incline; a force exerted parallel to the incline doesn't tend to bend or compress it.
• If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
We know that Fnet = m a, so we can see that .26N = 5kg *a, or a = 5.2 m/s^2.
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30 min
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You're on the right track, but see the link to be sure of your details.
&#Please compare your solutions with the expanded discussion at the link
Solution
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