Examkrackers Question #308

[3838]

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A 75 kg man holds a 25 kg pole at a 60 degree angle as shown. He leans backward at a 60 degree angle so that his center of gravity is 0.5 m to the right of his feet. How long is the pole.

a.4m
b.6m
c.7m
d.8m

Here's my attempt.

I chose a fulcrum at the man's feet.

x=distance from fulcrum to center of pole
y=distance from fulcrum to man's center)

x+y=1/2 the size of the pole.
25x=75y, x/y=3, x=3y
3y+y =4y=1/2 the size of the pole, 8y=size of pole
b/c y=0.5m, 4m should be the size of the pole.

Of course, I completely ignored the angular component of it. I couldn't visualize the trigonometry. I kind of simplified the length along the pole as the length of the lever, whereas the examkrackers book considered the distance from the fulcrum as the base of a triangle. I am really not sure how I was supposed to visualize this. Can anyone explain how I should conceptualize this problem? Obviously, there is something lacking in my understanding of a lever. For example, does the lever system that you choose always have to be horizontal? How do you handle angling such as in this case. I'm sorry if this comes across as a stupid/naw-duh question. I really just couldn't match this up with my concept of a lever very well. Something is obviously missing in my base of knowledge...😕

I want to understand how the problem was solved, of course, but I'm looking for an answer that will expand my understanding and intuition of a lever, that will help me visualize the lever in other situations so that I can apply the equations and concepts intuitively. Anyone?

 

[BB8730]

Is 4m the correct answer? I just did it quick and got 8m. If my answer is right I'll post my solution.

 

[3838]

Is 4m the correct answer? I just did it quick and got 8m. If my answer is right I'll post my solution.

8m is right! I'd like your solution, but can you also tell me how you visualized this as a lever? Because It was really hard for me to actually even treat this as a lever. I'm just really bad at rotational equilibrium. I missed like all of the last 1/2 of that section. I thought I understood it, but I find it really hard to "see" the levers sometimes. Like, where the fulcrum is, where the load is, and where the force is. any advice?

 

[BB8730]

Okay here's my solution. I didn't treat this as a lever or angular problem at all - just a center of gravity (mass) problem with a few twists to it.

So, first thing's first - the only thing touching the ground is the man's feet, so you know that his feet are at the center of gravity for the man + the pole. You don't know where the center of gravity of the pole is (because you don't know the length), so let's start with the man. The man is 75kg and his center of gravity is 0.5m to the right of his feet. If the pole weighed 75kg, its center of gravity would be 0.5m to the left of the man to balance it out. BUT the pole only weighs 25kg so we need to do some calculating. M1X1 = M2X2 is the equation to use here (mass * distance from center of gravity). (25kg)(X1) = (75kg)(0.5m). Solve for X and you get 1.5m. Now you know that the center of gravity for the pole is 1.5m to the left of the man's feet. As you can see from the drawing, the pole extends to 0.5m to the right of the man's feet, so the center of gravity for the pole is 2m to the left of the farthest right edge of the pole. In other words, 2m is half of the horizontal length of the pole. Some quick trig (cos(60) = adj/hyp, 4m = adj length) should give you 8m as the length of the pole.

Let me know if you have any questions about this solution.