Power by Motor on Pulley System

[k_squared]

Homework Statement

Homework Equations

P = T*V
F=ma
3. The Attempt at a Solution [/B]

My approach was this: Using the pin at C as a datum, I figured that DB=CD and thus 2CD + AC = l. Thus I figured that at 3 ft/s, A was rising 3/2 feet per second.

However, I also figured the it would just be (3/2=v)(F=force required to keep box from accelerating, as the speed is constant.)I'm apparently wrong about this. The book gives the equation:
T₁ - 100 = 0
100 + 100 -T₂ = 0

...I take it T₂ is on the motor, but, I'm not sure. Could someone please give me a hint about the tensions? These pulley problems have been killing me even where everything else is easy!

 

[Mister T]

T₁ - 100 = 0
100 + 100 -T₂ = 0

Evidently T₁ is the tension in the piece of rope between A and C.

T₂ is the tension in the piece of rope between B and E.

So the motor is exerting a force of 200 lb.

My approach was this: Using the pin at C as a datum, I figured that DB=CD and thus 2CD + AC = l. Thus I figured that at 3 ft/s, A was rising 3/2 feet per second.

I don't follow. I think you meant BC=BD?

Anyway, the work done by the motor on the rope has to equal the work done on the load by the rope.

 

[k_squared]

Why is the tension 200... is is because the pulling is from both sides?

 

[Mister T]

Why is the tension 200... is is because the pulling is from both sides?

When you solve the equation

100 + 100 -T₂ = 0

for T₂, you get 200 lb.

It's because there are two ropes pulling upward on the pulley B, and each rope has a tension of 100 lb.

Does the statement I made about the work done help you?

 

[k_squared]

Ahh... I understand. The tension in the rope attached to the load is uniformly 100 pounds, and the motor is pulling against *two* such segments. That kinda makes sense. I have no problem whatsoever computing the work power on systems without pulleys... I seem generically bad at getting the tension, though.

So technically... no, the last thing you said was the only thing that didn't help me! So basically, this set up is backwards, because the load has mechanical advantage on the motor?

Would that mean that if the load and the motor were switched, the motor would only have to develop 50 lbs of tension in the rope to be in equilibrium?

 

[Mister T]

The statement I referred to was an attempt to get you to understand the speed of the load. You have it wrong in your attempted solution. I also asked you about a typo you may have made in labeling the rope segments?

 

[k_squared]

Yes, I meant BC=BD... I'm still not sure I understand the speed of the load, I'm going to review that section if I can find it.

 

[Mister T]

The work done by the motor on the rope has to equal the work done on the load by the rope. Work equals force times distance. If the force on rope BE is two times the force on rope AC then what does that tell you about the distance traveled by each section of rope?

 

[k_squared]

...BE travels half the distance as AC?

 

[Mister T]

Right, so that tells you the relationship between the speeds. You had it backwards originally!