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HMCCSUF
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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) let's 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):
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 let's 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
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) let's 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 let's 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