# Kinematics and Instantaneous Changes

#### HMCCSUF

Here is a simplified dynamics test problem that a professor and I have been discussing, and I would like another opinion (or opinions) on the issues at hand. I'll state the problem per the template, and I'll also give the jist of the arguments on either side. Any insight would be great.

1. Problem Statement:A safe with a weight of 200 lbf is being hoisted by two people with a rope and pulley arrangement at an acceleration of 2 ft/sec2 upwards (note: pulley ratio 2:1 eq. force on safe to applied force from people). Person A weighs 90 lbf. Determine the rate of acceleration of the safe if person B (weight not given) lets go of the rope at t=0 and leaves person A holding the rope by himself.

2. The applicable equations:
Neglecting the weight of the pulleys and rope, and assuming uniform gravity, the applicable equations are as follows (just use the force on the safe as 2:1, it's not the crux of the issue):
F = m * a​
Differential kinematic equations apply​

3. The Attempt at a Solution :
a. First solution:
Neglect t<0 conditions.
At t=0+, the FBD for the safe is 180 lbf up, and 200 lbf down.
This implies that 20 lbf=m*anet, which is down for the safe, and up for person A.
m, then, is the mass of the system, or 90 lbf + 200 lbf = 290 lbf = W, meaning m=290/32.2 = 9 slugs
Solving for the net acceleration, we obtain the system's acceleration of 20/9.0, or 2.22 ft/sec2, down for the safe.

b. Second solution:
The initial tension is governed by 2T-200lbf=m*a(safe), where T is the tension, and 2T is the total upwards force due to the pulley ratio. This implies that the initial tension in the rope is 106.2 lbf. The force balance for t=0+, then, is concentrated on person A. Here, we move to person A because we know his weight. Thus, the F=ma equation becomes 106.2 lbf (the tension in the rope) - 90 lbf (weight of person A) = 90/32.2 (=mass)*a(person A). Solving for a(person A), we obtain 5.8 ft/sec2. This implies that the acceleration of the safe is -2.9 ft/sec2, again in the opposite direction, down.

My question is who is right? One argument (b) holds that the tension remains the same as time goes from negative to positive. Argument (a) is essentially that both the tension and the acceleration change instantaneously. Since F=ma, and m changes instantaneously (one person lets go of the rope), does the tension change, the acceleration change, or both change instantaneously?

Some things to consider might be "jerk", or da/dt, and/or the rate of change in forces through a medium.

Love to hear some feedback!

Thanks,
HMCCSUF

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#### turin

Homework Helper
Differential kinematic equations apply
I don't think that these are needed.

At t=0+, the FBD for the safe is 180 lbf up, and 200 lbf down.
This implies that 20 lbf=m*anet, which is down for the safe, ...
Yep.

m, then, is the mass of the system, or 90 lbf + 200 lbf = 290 lbf = W,
Nope. m is the mass of the object in the fbd. The fbd allows you to determine the net F acting on the object, and then Newton's second law tells you how to relate this net F to the mass and acceleration of that object.

The initial tension is governed by 2T-200lbf=m*a(safe), where T is the tension, and 2T is the total upwards force due to the pulley ratio.
Yep.

This implies that the initial tension in the rope is 106.2 lbf.
I don't know how you got that. I'm not familiar with the British system, and I don't care enough to familiarize myself with it. If you show your equation for this (independent of the unit system), then I can tell you if it is correct.

The force balance for t=0+, then, is concentrated on person A.
I don't know what this means. Again, if you show and equation, then I can tell you if it is correct.

Thus, the F=ma equation becomes 106.2 lbf (the tension in the rope) - 90 lbf (weight of person A) = 90/32.2 (=mass)*a(person A). Solving for a(person A), we obtain 5.8 ft/sec2. This implies that the acceleration of the safe is -2.9 ft/sec2, again in the opposite direction, down.
Too many numbers, without showing the physics formulae. Again, if you show the equations ...

Argument (a) is essentially that both the tension and the acceleration change instantaneously.
This is an approximation that also I would use.

Since F=ma, and m changes instantaneously (one person lets go of the rope), does the tension change, the acceleration change, or both change instantaneously?
You need to be more specific. Mass of what? Acceleration of what?

Some things to consider might be "jerk", or da/dt, and/or the rate of change in forces through a medium.
Newtonian mechanics is essentiall second order in time, so jerk is not something that is constrained by the theory. Of course, in realistic materials, there is a finite time for a tension to be transmitted through a cord, and then there is a finite time during which the tension oscillates and settles to its final value. If that is essential to the problem, then it just got a mess-load more complicated than I want to deal with. (Not to mention that you are going to need a mess-load more of information.)

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#### HMCCSUF

Nope. m is the mass of the object in the fbd. The fbd allows you to determine the net F acting on the object, and then Newton's second law tells you how to relate this net F to the mass and acceleration of that object.
That would be a fundamental concept and a goof on my part -- thanks for straightening that out.

At steady state, the acceleration in the safe would simply be governed by Ftotal=msafe* asafe, correct? Since the mass of the safe is 200 lbf/32.2 ft/sec2, this would translate to (200 lbf-180 lbf) = 6.21 slugs * asafe. This should be right -- if not, I've been thinking about this problem for way too long and have lost lucidity.

I don't know how you got that. I'm not familiar with the British system, and I don't care enough to familiarize myself with it.
Can't say I blame you. This calc was done by saying the safe is accelerating up at 2 ft/sec2, which is 1.06*g. Since the safe "weighs" 200 lbf, the force required to do this is 212 lbf. However, the mechanical advantage is 2, so the tension drops back down to 106 lbf. That said, let's ignore that calculation since it's not really crucial to my confusion about the instantaneous changes.

You need to be more specific. Mass of what? Acceleration of what?
Good point. I meant that the mass of the people pulling on the rope changes; this isn't really a mass as we've analyzed it, but a force imparted to the safe.

So, if person B lets go of the rope, does the acceleration of the safe change instantaneously? If person B lets go of the rope, does the tension in the rope change instantaneously?

Or, pointedly, can the acceleration of the safe change instantaneously without the tension in the rope changing instantaneously?

I don't want to worry about specifics, as you discuss below. The complications added by any SHM are beyond the scope of this class.

Newtonian mechanics is essentiall second order in time, so jerk is not something that is constrained by the theory. Of course, in realistic materials, there is a finite time for a tension to be transmitted through a cord, and then there is a finite time during which the tension oscillates and settles to its final value. If that is essential to the problem, then it just got a mess-load more complicated than I want to deal with. (Not to mention that you are going to need a mess-load more of information.)
OK, sounds good. I tried to use the "definition" of jerk to show how a differential force should result in a differential change in acceleration if the mass in question remains the same. I think this is true, but it could conflict with the constraints of real-world applications, like the time needed for the rope to transmit the tension.

Thanks for your take on this. And to think I almost posted on Yahoo answers . . .

#### turin

Homework Helper
I'm sorry, I need to straighten out an error that I made in my previous post. The net force on the safe is NOT (msafe-2mA)g downwards, as I mistakenly agreed to. The reason is that T≠mAg, where T is the tension in the rope, because A is accelerating upwards. So, T not only has to counteract mAg, but also to accelerate A, which increases the amount of tension needed. So, my first "yep" should have been a "nope", with this explanation.

#### turin

Homework Helper
At steady state, the acceleration in the safe would simply be governed by Ftotal=msafe* asafe, correct?
I don't know what you mean by "Steady State" in this context. (Perhaps you are alluding to my comment about oscillations?) At any rate, that equation is always correct in Newtonian mechanics, assuming that Ftot is the force applied directly to the safe, and that the safe is perfectly rigid. However, determining Ftot may be quite complicated, and an alternative effective total force, e.g. from the force applied to the other end of the rope, would then only apply if the oscillations had died out.

I meant that the mass of the people pulling on the rope changes; this isn't really a mass as we've analyzed it, but a force imparted to the safe.
OK. I don't see anything wrong with that.

So, if person B lets go of the rope, does the acceleration of the safe change instantaneously?
That's the issue. If you just deal with the simplified approach that the force is transmitted through the rope instantaneously, then yes. However, if you want to deal with finite the elasticity of the rope, then no.

Or, pointedly, can the acceleration of the safe change instantaneously without the tension in the rope changing instantaneously?
I don't think that Newtonian mechanics prevent that, but in this situation, I don't think so.

And to think I almost posted on Yahoo answers . . .
Noooo ....

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