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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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cq_1_172
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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.
** **
cq_1_172
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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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
I have drawn a picture. It started with a regular x-y coordinate system with a line that
had an angle of 30 degrees from the origin. There is a 5 kg cart drawn at the origin
starting up the hill. Then I drew move lines with arrows--one on the incline and a
second perpendicular to this. I am thinking because I had the original perpendicular
lines, an angle of 30 degrees and then new perpendicular lines using this angle, that
the gravitational force has a 30 degree angle from the y axis and a 60 degree angle from
the x axis. But I don't really know.
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You don't mention a vector for the weight of the cart.
That's the very first force vector you need to draw. It is directed vertically downward.
What angle does that vector make with the positive x axis?
What is its magnitude?
What therefore are its components?
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The force vector that goes down vertically would make a 60 degree angle with the incline
and x axis. Since it's gravity, it would be 9.8 m/s^2. I did not multiply it by the
mass of the cart, although maybe I should. However, with 9.8 m/s^2, the x is 4.9 and
the y is 8.
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If the weight of the cart was 9.8 Newtons then the x and y components would have those
magnitudes, which is a good start. However that is not the weight, and the components
would not both be positive, so there's still a little ways to go.
First, the 5 kg cart has weight 49 Newtons.
Next, the vector does make a 60 degree angle with the x axis, which is good in itself.
But that is not its angle measured counterclockwise from the positive x axis.
The incline rises as you move to the right. so the positive x axis is up and to the
right. The 60 degree angle is therefore made between the weight vector and the negative
x axis.
Measuring the angle in the counterclockwise direction from the positive x axis we go
through 180 degrees to get to the negative x axis, then we go through your 60 degree
angle to get to the downward vertical weight vector. That makes 240 degrees.
What therefore are the components of the weight vector?
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####49 cos 240 deg = -24.5 for x
49 sin 240 deg = -42.4 for y ####
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Good.
That would be
49 N cos 240 deg = -24.5 N for x
49 N sin 240 deg = -42.4 N for y
It's always important to keep track of the units.
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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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
sine: the trigonometric function that is equal to the ratio of the side opposite a given
angle in a right angle to the hypotenuse. The sine would show the y component.
cosine: the trigonometric function that is equal to the ratio of the side adjacent to an
acute angle in a right triangle to the hypotenuse. The cosine would show the x
component.
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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 force against the cart by the incline would be upward. I just don't know if it's the
30 degree or 60 degree push.
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Whichever component of gravity is pushing into the incline, that's the one the incline pushes back against.
Which would that be, and what would be the push of 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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
This would be the force of gravity--9.8 m/s^2
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If the incline wasn't there the cart would accelerate at 9.8 m/s^2.
However the incline is there, and it prevents the cart from accelerating at this rate.
The weight, i.e., the force of gravity on the cart, is equivalent to the two components
you have found. One of those components acts in the x direction and another in the y
direction.
The y component is perpendicular to the incline and therefore pushes direction into the
incline, which (if it doesn't break) compresses and/or bends, resulting in an equal and
opposite force pushing back on the object. This force is called the normal force.
The y component of the weight and the normal force add up to 0.
This leaves only the x component of the weight to accelerated the cart down the incline.
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####The x component was -24.5. I hope I answered all of the unanswered questions on this####
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That would be -24.5 Newtons, and that would be the net force.
So the acceleration of the cart is
a = F_net / m = -24.5 N / (5 kg) = -4.9 m/s^2.
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*#&!
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No revision is necessary, though questions are still welcome.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
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Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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