Phy 201
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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: ->->->->->->->->->->->-> :
#$&* Wouldn’t gravity make a 90 degree angle with the positive x–axis since it’s acting straight vertically down? I think the x component of gravitational force will be greater. Since the incline is at 30 degrees, the x-axis would have to be greater than the y-axis.
The y axis is at right angles to the x axis; the entire coordinate system is rotations so that the x axis is directed up and to the right in the direction parallel to the incline.
• Using the definitions of the sine and cosine, find the components of the cart's weight parallel and perpendicular to the incline.
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* It’s been a long time since I used sine and cosine, but I think sin ‘theta = (opposite side/hypotenuse) and cos ‘theta = (adjacent side/hypotenuse). But I am drawing a complete blank on how to use the 30 degrees to determine the x and y sides….
• 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: ->->->->->->->->->->->-> :
#$&* I think the compression force must be perpendicular to the incline, since it has to work equal and opposite to the cart’s weight. I would guess the elastic force must be – 49 N. I say this because weight = (m * g) = 5kg * 9.8m/s^2 = 49 N.
• If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
#$&* Won’t acceleration be the acceleration due to gravity? However, I recall from other problems when g would be multiplied with the percentage of the slope. Then that is multiplied by the mass of the object, and that becomes the net force of the object.
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About 30 minutes...
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I have forgotten how to use sine and cosine. Its been a long time. And when the degree of the slope is given I'm not 100% sure how to find the weight. I know the mass and the acceleration due to gravity must must used, but I have a feeling I have to do something with that given 30 degrees...
&#Please compare your solutions with the expanded discussion at the link
Solution
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