Work Equation

[arlo]

I need a clarification on what the angle is in the work equation.

W = Fd cos (theta)

I know that the angle is suppose to be between force and displacement vector. I'm confused on when I need to use the angle when I calculate work. For example, AAMC Qpack physics #74 (I'm not sure I can post screenshot of the problem). They calculated work in the pulley by just W = Fd even though the diagram showed the string was pulled at an angle. Can someone break this down to clarify when we factor in the angle to the equation? Thanks.

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[theonlytycrane]

You want to multiply the component of the force in the direction of displacement. Sometimes you will need the angle for only a component of the force or sometimes you just use the entire force if it's in the same direction as displacement (in which case cos(0) = 1).

 

[aldol16]

Work is always equal to the component of the force that is applied in the direction of displacement multiplied by the displacement. This is what I hate about the AAMC course requirements - they only require non-calculus-based physics! So if you understand calculus-based physics, it should be fairly obvious what theta is because it's just a dot-product and a dot-product is a re-iteration of what something like work is. But for non-calculus-based physics, how they teach it is to draw a triangle.

Say you have a force that is being applied to move a box across a frictionless floor. The force of 2 N is applied at an angle of 45 degrees with the horizontal to move the box 4 m across the floor. The work is therefore the component of force in the horizontal direction multiplied by 4 m. So what's the component of force in the horizontal direction? Well, you have to decompose the force vector into vertical and horizontal components. Let's call the horizontal component x. So then according to the laws of right triangles, the cosine of 45 degrees must equal the adjacent side divided by the hypotenuse of the triangle. The hypotenuse is 2 N. The adjacent side measures x. Thus, cos(45 degrees) = x/2 and x = 2*cos(45). That's where you get the work equation.

 

[arlo]

Work is always equal to the component of the force that is applied in the direction of displacement multiplied by the displacement. This is what I hate about the AAMC course requirements - they only require non-calculus-based physics! So if you understand calculus-based physics, it should be fairly obvious what theta is because it's just a dot-product and a dot-product is a re-iteration of what something like work is. But for non-calculus-based physics, how they teach it is to draw a triangle.

Say you have a force that is being applied to move a box across a frictionless floor. The force of 2 N is applied at an angle of 45 degrees with the horizontal to move the box 4 m across the floor. The work is therefore the component of force in the horizontal direction multiplied by 4 m. So what's the component of force in the horizontal direction? Well, you have to decompose the force vector into vertical and horizontal components. Let's call the horizontal component x. So then according to the laws of right triangles, the cosine of 45 degrees must equal the adjacent side divided by the hypotenuse of the triangle. The hypotenuse is 2 N. The adjacent side measures x. Thus, cos(45 degrees) = x/2 and x = 2*cos(45). That's where you get the work equation.

What you said made sense. That's the way I would approach it to find work. In the qpack problem though, the AAMC explanation did not take the angle into consideration. The way I approached it by incorporating the angle into W = Fd cos (theta), I got W = (4 kg)(10 m/s^2) cos (30) x 5 m = ~177 J. The answer is 200 J which is the closest. I'm just confused why they didn't use the angle and if I'm wrong for using it. The reason I used the angle is because F makes 60 degrees with the ceiling to move 4 kg up 5 m.

 

[NextStepTutor_2]

I need a clarification on what the angle is in the work equation.

W = Fd cos (theta)

I know that the angle is suppose to be between force and displacement vector. I'm confused on when I need to use the angle when I calculate work. For example, AAMC Qpack physics #74 (I'm not sure I can post screenshot of the problem). They calculated work in the pulley by just W = Fd even though the diagram showed the string was pulled at an angle. Can someone break this down to clarify when we factor in the angle to the equation? Thanks.

You do not need any calculus on the MCAT so if you have never had it (many pre meds have not, at least not in a physics class, you're not missing out on anything too exciting, I promise ) do not look into it or study it for the MCAT as it will be a waste of your precious study time. Most of my classmates in medical school had taken just 1 semester of calculus, and almost all had forgotten it. You won't need it for medical school but if you do know it, great as it can help with grasping some of these concepts and equations but, even after 6 semesters of calculus in undergrad, I never needed it on the MCAT or in medical school.

This problems is straightforward. The work done by the pulley better be equal to the work done on the object. Using a pulley or any mechanical device means I can utilize a lower force over a longer distance to perform the same work. What must hold for all situations is that the work done by the pulley must be equal to the work required to lift the object, that is it.

Work by pulley = F(sin(60))Drope = Work needed to lift 4 kg mass 5 meters into the air = 4(g)5 = 4x10x5 = 200 Joules

The angle here is irrelevant to the question they asked since we already have the information we need. if they had asked what F is needed on the pulley, or what distance or rope you will need to pull, then the angle would matter.

Long story short, whether you use a lever, a ramp, or a pulley, the work required to lift mass M to height H will always be MgH. The AAMC will often put unnecessary information in a discrete problem (and always in a passage) JUST TO DISTRACT YOU. This is why science understanding (with the MCAT scope in mind) trumps memorization every time.

Is this simple? You bet, but so is the physics on the MCAT.

Hope this helps, good luck!