Conservation of energy: Crane raising a mass

[WherE mE weeD]

Homework Statement

Mechanical crane raises 225kg at a rate of 0.031m/s^2. from a rest to a speed of 0.5m/s over a distance of 4m. Frictional resistance is 112N.
m1=225kg
a1=0.031m/s^2
u1=0m/s
v1=0.5m/s
s=4m
Fr=112N
A. Work input from the motor
B. Tension in the lifting cable
C. Max power developed by the motor.
D. Using D'Alembert's principle, solve a) b) c)

Homework Equations

Wd = ΔPE + ΔKE + Wf (Wf = Work done to overcome friction)
Fnet = ma
Wd = Fs
Weight =m(a+g)
PE = mgh
KE = mv^2/2
D'Alembert's principle; ΣF = 0
Mechanical Power P = Energy used/Time taken P = FV
v=u+at
t=(v-u)/a
Wd = Work done

The Attempt at a Solution

PEi = mgh = 225x9.81x4 = 8829J (incorrect)
PEi = mgh = 225x9.81x0 = 0J
PEf = 0 (as the energy is transformed to KE) (incorrect)
PEf = mgh = 225x9.81x4 = 8829J
KEi = 0 (starts from a rest)
KEf = mv^2/2 = 225(0.5x0.5)/2 = 28.13J
Wd to overcome friction during motion: Wd = Fs = 112x4 = 448J
Wd (motor) = (PEf - PEi) + (KEf - KEi) + Wf = (8829 - 0)+(28.13-0)+448 = 8352.87
Therefore the engine Wd would be 8352.87J

B) Tension in the rope: Tension = weight + Fnet = (ma)+(mg)
In this case the acceleration of the mass is opposite the acceleration of gravity therefore the formula would be. This is assuming Tension during the acceleration of the mass upwards.
Tension =m(a+g)
=(225x0.031)+(225x9.81)
=2214.23N force of tension on the rope

c.)
P=Fv
=(rope tension + Ff)0.031 = (2214.23+112)0.031 = 72.11W

D'Alemberts principle states that the total forces must add upto 0. I am not sure how I could show this.

To me the power output seems to low and I feel like I am missing something on energy when calculating the work done, any help would be mucho appreciated sorry about the math grammar.

 

[TomHart]

PEi = mgh = 225x9.81x4 = 8829J
PEf = 0 (as the energy is transformed to KE)

Didn't the problem state that the crane "raises" the mass? Then why would the potential energy be less when it is higher?

 

[WherE mE weeD]

Didn't the problem state that the crane "raises" the mass? Then why would the potential energy be less when it is higher?

It does. I see your point. So I would calculate height as zero for initial PE and PEf would be calculated at the final height destination of the mass in this situation.

 

[WherE mE weeD]

Thanks for the prompt reply Tom

 

[TomHart]

Yes, I would start out with initial potential energy as 0. That would be the simplest, I think.

 

[WherE mE weeD]

Homework Statement

Mechanical crane raises 225kg at a rate of 0.031m/s^2. from a rest velocity 0m/s to a final velocity of 0.5m/s over a distance of 4m. Frictional resistance is 112N.
m1=225kg
a1=0.031m/s^2
u1=0m/s
v1=0.5m/s
s=4m
Fr=112N

B. Tension in the lifting cable

B) Tension in the rope: Tension = weight + Fnet = (ma)+(mg)
In this case the acceleration of the mass is opposite the acceleration of gravity therefore the formula would be. To find the resultant Tension during the acceleration of the mass upwards.
Tension =m(a+g)
=(225x0.031)+(225x9.81)
=2214.23N force of tension on the rope

Can anybody see where I am going wrong in my calculations to find tension in the lifting rope. The question is very vague as it doesn't mention if the tension is to be found while at rest or during acceleration.

Thanks in advance for help.

 

[TomHart]

Isn't there a friction force also?

 

[WherE mE weeD]

Yeah the friction force is 112N.
Ahh yes so that would be added to the make the total Fnet.
Tension = weight + Fnet = (ma)+(mg)+ Fr
Tension =(225x0.031)+(225x9.81)+112 = 2326.23N

 

[WherE mE weeD]

Would this look about right for the formula's
Applied force = (Weight+Ff)+(ma)
Tension = (Weight+Ff)+applied force

Sorry about the bad drawing and hand writing lol

 

[TomHart]

As I rethink this problem, I think I am confused by 2 things.
1) Where does the friction force act? Is the friction force acting on the mass? Let's assume for now that it does.
2) I'm kind of confused by your force definitions. Your diagram shows three forces acting on the mass: friction force, weight force, and force applied. So I assume that the force applied is acting on the mass at the point where the cable attaches to the mass. Is that true? If so, that makes sense to me. So, then where does tension come in? Is that a force acting on the mass, or is that a force somewhere else?

Maybe the friction force was not intended to be acting on the mass. Do you know for sure?

 

[WherE mE weeD]

As I rethink this problem, I think I am confused by 2 things.
1) Where does the friction force act? Is the friction force acting on the mass? Let's assume for now that it does.
2) I'm kind of confused by your force definitions. Your diagram shows three forces acting on the mass: friction force, weight force, and force applied. So I assume that the force applied is acting on the mass at the point where the cable attaches to the mass. Is that true? If so, that makes sense to me. So, then where does tension come in? Is that a force acting on the mass, or is that a force somewhere else?

Maybe the friction force was not intended to be acting on the mass. Do you know for sure?

Sorry I have not given all the information required Tom I will give the answers now.
1.) Frictional force is resistance to the motion of the mass.
2.)Yes, The applied force is the force acting at the point between the cable and the mass which in effect will raise the mass at the specified acceleration.

Tension is the force in the lifting cable I imagined Tension as all the forces acting on the rope both vertically and horizontally?

 

[WherE mE weeD]

Sorry for throwing the question out there without the required information.

 

[TomHart]

So if the friction is acting on the mass, then there is no other friction that the cable experiences, or in a pulley that the cable moves around. So then the tension in the cable (at any point in the cable) is the same as the applied force. So then the motor has to do work according to that tension. So I think it is really unnecessary that you have a force called "tension" and another force called "force applied". In this problem, I think my preference would be to use "Tension" rather than "force applied". That's just my preference though.

Now let's say you had a pulley that had friction. In that case, the tension in the rope would be different on one side of the pulley as it would on the other side.

 

[WherE mE weeD]

Ok so take out the force applied and stick with the tension force. And that would be the result of the tension in the rope between the mass and the pulley?

A question I have is the equation F=ma is used to find the force required to pulley the mass upwards? I was just wondering as it is such a small force compared to the forces pushing against the mass.

 

[TomHart]

F = ma should actually be ΣF = ma. The sum of the forces is proportional to the acceleration. Let's forget the friction force for a moment. Let's say you have an enormously large weight (w) connected to a cable. Now if the tension on that cable is also enormously large, but just slightly larger than the weight, then the sum of the forces will be very small. So you will have a very small net force acting on a very large mass, which results in a very small acceleration. Does that make sense?

Here are some numbers as an example. Mass = 1000 kg. So if g = 10 m/s^2, then the weight is 10,000 N. Now if the tension in the cable happens to be 10,001 N, then, because the tension force is greater than the weight force, the object will have an acceleration upward. However, because the difference is very small, then the acceleration will also be very small. ΣF = 10001 - 10000 = 1 N.
And ΣF = ma so a = ΣF/m = 1/1000 = 0.001 m/s^2.

 

[WherE mE weeD]

Thats makes sense so ΣF is the net force which is the force proportional to the acceleration of the mass.
Tension in the rope is ΣF + weight this will show whether the mass is being lifted or lowered and by how much force.
ΣF = ma = 225kg x 0.031m/s^2 = 6.89N
Tension = ΣF + w = 6.89N + 2207.25N = 2214.14N

Thanks Tom very helpful the way you describe these situations makes it very easy to understand for myself.

So the resistance force to movement of the mass this would be also acting on the rope through the mass and could be added into the tension equation as:
Tension = ΣF + w + Ff = 6.89N + 2207.25N + 112N = 2326.14N

 

[TomHart]

So the resistance force to movement of the mass this would be also acting on the rope through the mass and could be added into the tension equation as:
Tension = ΣF + w + Ff = 6.89N + 2207.25N + 112N = 2326.14N

Correct.

 

[WherE mE weeD]

D. Using D'Alembert's principle, solve a) b) c)

The Attempt at a Solution

PEi = mgh = 225x9.81x0 = 0J
PEf = mgh = 225x9.81x4 = 8829J
KEi = 0 (starts from a rest)
KEf = mv^2/2 = 225(0.5x0.5)/2 = 28.13J
Wd to overcome friction during motion: Wd = Fs = 112x4 = 448J
Wd (motor) = (PEf - PEi) + (KEf - KEi) + Wf = (8829 - 0)+(28.13-0)+448 = 9305.13J
Therefore the engine Wd would be 9305.13J

B) Tension in the rope
ΣF = ma = 225kg x 0.031m/s^2 = 6.89N
Tension = ΣF + w = 6.89N + 2207.25N = 2214.14N
Tension = ΣF + w + Ff = 6.89N + 2207.25N + 112N = 2326.14N

c.)
P=Fv
=(rope tension)0.6 = 2326.14 x 0.6 = 1395.68W

D'Alemberts principle states that the sum of forces must add upto 0 due to the force acting in opposite equal equillibrium.
ΣF-ma=0.

A.) I'm not sure how the equation ΣF-ma=0 could apply to a problem using energy. my closest guess which is basically using the conservation of energy which is the energy input minus the energy output equals 0. This is really confusing for me as I always thought D'Alembert's principle referred to forces only.

B.) I can easily prove ΣF-ma=0
using the previously calculated numbers.
6.89N - (225kg x 0.031m/s^2) = 0

C.) Again I am confused how proving the forces add upto 0 can help to find the power output. I feel like I understand D'Alemberts principle and that it is used to convert a dynamic problem to a static by proving their is an equal and opposite force in equilibrium to all forces.

Thanks again for any help and big up Tom who has been giving a helping hand throughout my problem solving.