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# Components of a Force - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We're talking about the components of a force, and here I have kind of a fun problem. Sarah Stickman of Stickman and Sons Movers is holding a safe (loaded on a dolley) on a 15 degree incline while her brother takes a coffee break. So here in the picture I've got Sarah holding onto a safe, here if the 15 degree incline and here is her brother drinking coffee.

If she is exerting 100 pounds of force, how much does the safe weigh? So imagine that she is applying a force factor of 100 pounds that is directed in this direction parallel to the plane, the weight is directed directly downward, we have to find the magnitude of that force.

So I've enlarged this picture, and I've included the vectors I just mentioned. This is the vector representing the weight, this is her 100 pounds of force. Now here we'll have to use Newton's third Law, 'Every Reaction has an equal and opposite Reaction'.

This 100 pound force that Sarah is applying has an equal and opposite force that's applied to her from the box, from the safe. And this force is the component of weight that's parallel to the surface of the plane. So I can draw it down here as well and so I know this is going to be 100 pounds.

Now the vertical component I don't know what the magnitude is, but I can certainly draw it. It's going to look like this. Now this forms a right triangle, because these are components, they're perpendicular to one another. So I have a right triangle. Now remembering that the angle of incline is 15 degrees. I know that this angle, this acute angle here is also 15 degrees. So the key to finding the value of the weight is using right triangle trigonometry.

The sine of 15 degrees is going to be 100 pounds divided by the magnitude of the weight. Sine of 15 degrees is 100 pounds divided by the magnitude of w. So to find the weight, I multiply both sides by the magnitude of w, and I get w times sine of 15 equals 100 pounds, then I divide both sides by the sine of 15 degrees. And I can easily calculate this on my calculator, so let me do that now. 100 divided by sine 15 degrees, I get 386.4 pounds, and that's the magnitude of the weight.

We know that the weight is directed downwards, so we don't need to worry about direction so much. But we found that the safe, if she's pushing with 100 pounds of force and the angle is 15 degrees, the safe is going to weigh 386 pounds.

Now it makes sense that if the angle had been bigger, the components of force parallel to the plane would be greater and she would have to push harder to keep the safe in place. And if the angle were smaller, she'd have to push less hard. But the key to finding the magnitude of the weight, is setting up a right triangle, and solving for the hypotenuse.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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