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.
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Sketch this situation with the cart on the incline and the x axis 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.
answer/question/discussion: x = mg(cos 30) = .5 kg * 9.8 m/s^2 * .87 = 21.3 N, y = mg(sin 30) = .5 kg * 9.8 m/s^2 * .5 = 24.5 N
The gravitational force does not add at an angle of 30 degrees with the positive x axis, so using cos(30 deg) for the x component and sin(30 deg) for the y component is not appropriate.
The x axis is parallel to the incline. What therefore is the angle of the gravitational force, as measured counterclockwise from the positive x-axis?
This is the angle you need to use to calculate the components of the gravitational force.
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Using the definitions of the sine and cosine, find are the components of the cart's weight parallel and perpendicular to the incline.
answer/question/discussion: x = mg(cos 30) = .5 kg * 9.8 m/s^2 * .87 = 21.3 N, y = mg(sin 30) = .5 kg * 9.8 m/s^2 * .5 = 24.5 N
see my previous note, which is more appropriate to this question than the preceding. In the preceding question you were asked to base your estimate on the sketch, not on the calculation. Make sure you know how the sketch corresponds to the situation, as well as the calculation.
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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 elastic force should be vertical and must be equal to the vertical force of the 'mass of the car' * gravity.
When an incline is bent by a weight resting on it, the bend 'pushes back' elastically in a direction which is perpendicular to the incline. Unless the incline is vertical, this elastic reaction will not be vertical.
A compressive force on a rigid incline acts in a similar manner, with the compressive reaction being perpendicular to the incline.
An object on an incline can certainly experience a vertical upward force as a result of being on the incline. However any part of that vertical force which is not perpendicular to the incline is due either to friction or to some other mechanism (e.g., some attaching mechanism); if friction is not present and there is no such mechanism the object will slide down the incline, and the elastic reaction of the incline will not hinder its acceleration.
The elastic or compressive reaction of the incline is called the normal force, and as indicated in this note is perpendicular to the incline.
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If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: If the car is at rest, would the acceleration be zero, or would it be m * g * h?
The component of the gravitational force perpendicular to the incline is countered by the normal force (see preceding note).
The component of the gravitational force parallel to the incline is not countered in any way by the normal force. In the absence of friction or any other force, the net force on the object will therefore be equal to this component of the gravitational force.
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20 minutes
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This is a challenging situation, which once understood provides a key to many other situations.
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