Physics Problems

<p>These are two physics problems that we had on a test today, and I just can't figure out how to approach them.</p>

<p>A weightlifter who has a mass of 56 kg lifts a barbell that has a mass of 175 kg (I think those were the numbers...and yes, his mass was something small). Find the force exerted on each of his feet at the point where he is lifting the barbell at an acceleration of 1.0 m/s^2.</p>

<p>(No clue how to approach this problem. I said that the force exerted on his two feet combined was equal to: (175 kg)(9.8 m/s^2) + (175 kg)(1.0 m/s^2), because that's how much was necessary to overcome the gravity of the dumbbell.)</p>

<p>A wooden box of mass 10 kg is on an incline of 25 degrees. It is connected to a rope which is strung over a pulley at the top of the incline. On the other end is a hanging bucket of mass 2 kg. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.36. Find the acceleration of the block. </p>

<p>(In order to solve the problem, you need to find the friction...don't you first need to know whether it's moving or not, in order to know which coefficient to use?)</p>


<p>for the second problem. If it's accelerating, then it has to be moving so you must use the coefficient of kinetic friction. You can use the whole body method right here (where you combine the mass of the bucket and the block) since the tension in the rope will cancel out.</p>

<p>if you componentize gravity, the net force on the system is:</p>

<p>(Force of gravity on block) * sin(25) + Friction - Force of gravity on bucket</p>

<p>force of friction = uN = u(10)(9.8)(cos(25))</p>

<p>substitute numbers:</p>

<p>Fnet = (10)(9.8)sin(25) + .36(10)(9.8)(cos(25)) - (2)(9.8) = 53.79</p>

<p>set that equal to ma (remember m in this system is the combined mass of the block and the bucket)</p>

<p>53.79 = (10+2)(a)
a = 4.83 m/s^2</p>

<p>i think your part a is almost right except for the fact that you need to factor in his own weight.</p>

<p>for the first question:
you missed out the weightlifter's own weight, didn't you?</p>

<p>for the second:
yea,good point. But if the question is asking abotu acceleration, i would assume the block were already moving. After the box is moving - and so, accelerating - the coeff of static friction is irrelevant. Maybe that's the trick?</p>

<p>Hey everyone! Wow, thanks for the quick responses...just a few comments I would like to add...</p>

<p>Yeah, I figured that the weight of the weightlifter needed to be factored in. Thanks for that!</p>

<p>Also, I didn't copy the second question correctly. The question actually asked: "Determine whether the block stays at rest or accelerates. If it accelerates, determine the magnitude and direction of the acceleration."</p>

<p>ok, what you do for the second problem to determine if it stays at rest of accelerates is first use force of static friction, see if the forces balance (which they don't)... so that means it does accelerate (up the ramp)</p>

<p>Hmm...athlon, you're absolutely right. That's what I thought of doing. But is there another way of finding friction, independent of testing "trial and error" to see if the box moves? I ask this because the problem I just mentioned was actually part of a multi-step free response question. This was Part B. Part A asks for the force of friction. It just doesn't seem neat to be testing by trial and error (especially since you aren't asked to find the acceleration until you get to Part B).</p>

<p>aigh sorry athlon - screen didn't show your post. didn't repeat your post on purpose.</p>

<p>LOL, on the actual test, I actually used ma for the acceleration of the bucket, Ma for the acceleration of the box, and then mg sin 25 for the box. Grr...hopefully my teacher will realize this and have mercy on me............</p>

<p>Sorry to bump up this old post, but.........</p>

<p>How do you know which way friction is going? In athlon's post, friction is added to mg sin 25. I don't quite understand. Doesn't friction oppose the direction of motion?</p>

<p>friction is the pulling box downhill, and so is the parallel component of gravity</p>