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How to tell when to use Cos vs sin in physics

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data

    The first problem was " A 50 N crate is pulled up a 5m inclined plan by a worker at constant velocity. If the plane is inclined at an angle of 37degrees to the horizontal and there exists a constant frictional force of 10N between the crate and the surface, what is the force applied by the worker?"
    I knew that I could find the frictional force coefficient by 10=(coefficient)(mass)(gravity) =>
    10=(coefficient)(50N)(9.8) => coefficient =.02
    and then using the equation (coefficient)(mass)(gravity)(distance) I could get the force applied.

    The next problem was "A 40 N crate is pulled up a 5m inclined plan by a worker at constant velocity. If the plane is inclined at an angle of 37degrees to the horizontal and there exists a constant frictional force of 10N between the crate and the surface, what is the net change in potential energy?"
    I thought I would use (mass)(distance).

    With the first problem the distance would be cos(37) and in the second its sin(37)

    Why is that?


    2. Relevant equations
    (friction coefficient)(mass)(gravity)(distance)

    3. The attempt at a solution
    For the first, (50)(5cos(37))(9.8)(.02)=40N
    For the second, (5sin37)(40N)=120J
     
  2. jcsd
  3. Apr 9, 2015 #2

    collinsmark

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    Here is some general advice, not necessarily just for the examples you give.

    First, draw a free body diagram (FBD). Get in the habit of drawing them. They are not wasted effort; they actually do help.

    As part of the FBD, you'll notice that many things, forces, velocities, distances, etc., can be rearranged (rearranging the same type of things) into right triangles.

    Finally, commit these relationships to memory. There's not too many things that you need to remember in physics, but the below relationships are of the things you must remember (for right triangles):

    [tex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/tex]

    [tex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/tex]

    [tex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/tex]
     
  4. Apr 9, 2015 #3
    Yes, and don't forget to label your x and y axes. It might be easier to make your x-axis parallel to surface of the plane.
     
  5. Apr 9, 2015 #4
    I am still confused how to know which one to use. I drew the FBD for both problems and now I am staring at the charts trying to figure out why sin was used and why cos was used, in order to know which to use next time..
     
  6. Apr 9, 2015 #5

    collinsmark

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    Once you form a right triangle, it will have three vertices. One vertex, of course, will be a right angle. Then you must determine which of the other two vertices correspond to [itex] \theta [/itex]. That will depend on the problem you are working on. If the answer is not obvious from the problem statement's wording, you should be able to figure out which vertex is [itex] \theta [/itex] using geometry.

    From there, it should be easy to label which of the three sides of the triangle is the hypotenuse.

    The side opposite [itex] \theta [/itex] is the "opposite" side.

    The side of the triangle that meets up with the hypotenuse at the [itex] \theta [/itex] vertex (not the hypotenuse itself, but the other side that meets at [itex] \theta [/itex]) is the "adjacent."
     
  7. Apr 9, 2015 #6

    SammyS

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    sOH-cAH-tOA

    SOHCAHTOA
     
  8. Apr 9, 2015 #7
    Draw your object on the incline and label all the forces acting on it. It looks like you have 4. Start there and if possible, post a diagram so I can make some comments.
     
  9. Apr 9, 2015 #8

    collinsmark

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    By the way, I just went through your solutions, and I don't think they are quite right.

    In both problems, nothing is accelerating (the crate is at a constant velocity), so that makes things easier.

    Part A)
    In the first problem, let's start by ignoring friction for the moment. We'll come back to it. There are three forces acting on the crate (again, ignoring friction for now), its weight, the normal force, and worker's force. Knowing that the surface is always perpendicular to the normal, a right angle is formed. Put these three forces into a right triangle. Which vertex is [itex] \theta [/itex]? Solve for the force of the worker, ignoring friction.

    Now let's make sure the 10 N frictional force gets included somehow.

    Now ask yourself, "which direction does the frictional force point?" How much more does the worker's force have to be such that all forces still add to zero (ensuring there is no acceleration)?

    Part B)
    [itex] W = \vec F \cdot \vec d [/itex], where [itex] \vec d [/itex] is displacement.

    It simplifies to simply W = Fd if the force and distance are parallel to each other. Otherwise it involves the projection of one onto the other (involving the cos function), where [itex] \vec F \cdot \vec d = Fd \cos \phi [/itex], where [itex] \phi [/itex] is the angle between the force and displacement.

    But ask yourself, "In this problem, is the force that worker applied to the crate in the same direction as the crate's displacement?" if so, there's no need for a trigonometric function; they're already parallel [if you're tying to find the work done by the worker].

    [Edit: --- sorry, I didn't read the problem statement carefully on part B. I'll follow up on next post. The above is right the work done by the worker, but not the change in potential energy.]
     
    Last edited: Apr 9, 2015
  10. Apr 9, 2015 #9

    collinsmark

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    Sorry, in my last post on Part B, I explained the work done by the worker, not the change in potential energy.

    Let me revise my advice on part B (again, sorry for the confusion).

    Again, let's start with [itex] W = \vec F \cdot \vec d [/itex].

    Of all the forces involved on the crate, only one of them involves a change in potential energy. Which is it? Which direction does it point?

    This incline forms a right triangle, with a hypotenuse of 5 m. It has a 37 deg angle with the horizontal. So which component of this 5 m displacment is parallel with the force that changes potential energy? What is the length of that component?

    The final answer is the product of the force that involves potential energy and the displacement component that is parallel to that force.
     
    Last edited: Apr 9, 2015
  11. Apr 9, 2015 #10

    collinsmark

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    And welcome to Physics Forums, by the way. :smile:
     
  12. Apr 10, 2015 #11

    haruspex

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    Smith96, there are several issues with your solution to A.
    The weight of the crate is given as 50N, not as that due to a mass of 50kg. So it is wrong to multiply by g.
    Secondly, the kinetic frictional force is given by the coefficient multiplied by the normal force. Since it is on a slope, the normal force will not be the weight of the crate, but something less. However, that didn't affect your answer because it was not necessary to find the coefficient, so getting the wrong value didn't hurt.

    For the purposes of the problem, all that matters is that you have equilibrium between four forces at known angles, the magnitudes of two are known, and you wish to find a third.
    There are two unknowns (forces) and two available equations (equilibrium in each of two directions). Since you only care about one of the unknown forces, the convenient approach is to consider the equation for a direction at right angles to the one you don't care about. In this case, that will be parallel to the slope.
    So, how to pick between sin and cos for the three remaining forces, mg, force of pull and force of friction?
    Force of pull and force of friction are easy since they are both parallel to the plane. That leaves gravity.
    The contribution of a force to a linear balance is always the cosine of the angle between the force and the direction of interest. (Linear as distinct from angular; for torques it's different.)
    So all you have to decide is whether the angle between mg and the slope is theta or its complement. A check I use is to think about the theta=0 case. If the angle of the slope to the horizontal were zero then gravity would have no component along the slope, so we must want sine.
     
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