Statics Problem: Crane with Suspended Ball | Help & Solutions | Mechanics

[Frisky90]

Homework Statement

Hello :) I am stuck on this mechanics problem. Any help would be hugely appreciated, because I am just starting with statics. Thank you in advance .

A diagram of the problem can be found here :http:////img97.imageshack.us/i/79504950.png

A small crane is used in demolition and swings a large rigid ball of mass M1, suspended on the lifting cable over a pulley of radius R, by rotating around the vertical axis zz with a steady velocity ω. The crane consists of a supporting uniform boom structure of mass M2 and length L inclined at an angle of θ to the horizontal plane. The boom is raised or lowered by the support cable that pulls horizontally and is attached at a distance of b from the pivot supporting the boom.

Note that the lifting cable is parallel to the boom.
The crane is shown in the static state and the ball is supported on the cable of length shown. For the dynamic conditions described, and for values:
h = 2 m; L = 2.5 m; R = 0.2 m; M1 = 250 kg; M2 = 200 kg;
ω = 1.0rad/s; θ = 60; b =1.5 m; r = 0.15 m;

Determine the following:

a) The radius R2 about which the ball moves about the zz axis.
b) The tension in the lifting cable.
c) The tension in the support cable.
d) The reactive pivot force at the base of the boom.
e) The axial force distribution along the boom.
f) The torque exerted by the lifting cable drum that has radius r.
You may assume that the mass of the cables and pulley are negligible and that the pulley is frictionless.

Homework Equations

Method of Joints,method of sections,equilibrium equations, distributed loads along a line(F=integral_L w dx)

The Attempt at a Solution

I did a free body diagram,including the boom, the lifting cable and the base.
F1=250kg*9.8=2450N( is this the tension in the lifting cable-Q.b?) , F2=200*cos60*9.8=980N ( ans. to Q.c ?)

 

[Colt 45 J]

Show your FBD and it's a lot easier to see where you went wrong?

 

[Frisky90]

Show your FBD and it's a lot easier to see where you went wrong?

Well, I've shown you my answers to Q.b and c. Are they correct?
Any suggestions how to continue?

 

[nvn]

Frisky90: Use g = 9.81 m/s^2, instead of 9.8. Try your answer for (b) again, using this value for g. By the way, please note the following international standard for writing units.

1. Always leave a space between a numeric value and its following unit symbol. E.g., 2452.5 N, not 2452.5N. See the international standard for writing units (ISO 31-0).

Your answer for (c) is incorrect. Try again.

 

[Frisky90]

Frisky90: Use g = 9.81 m/s^2, instead of 9.8. Try your answer for (b) again, using this value for g. By the way, please note the following international standard for writing units.

1. Always leave a space between a numeric value and its following unit symbol. E.g., 2452.5*N, not 2452.5N. See the international standard for writing units (ISO 31-0).

Your answer for (c) is incorrect. Try again.

Thank you for the response. Using 9.81, I got 2452.5 N . Is that the answer (Q.b) ?
Should I use the formula for loads distributed along a line for Q.C ?

 

[nvn]

Frisky90: Your answer for (b) is correct. Nice work. No, you do not need to use the formula for distributed loads for question (c), because the crane boom weight can be assumed to act as a point load downward at the crane boom midpoint for question (c) ... but not for questions (d) and (e). I.e., you would not need to consider the distributed boom force until questions (d) and (e).

 

[Frisky90]

Frisky90: Your answer for (b) is correct. Nice work. No, you do not need to use the formula for distributed loads for question (c), because the crane boom weight can be assumed to act as a point load downward at the crane boom midpoint for question (c) ... but not for questions (d) and (e). I.e., you would not need to consider the distributed boom force until questions (d) and (e).

Thx,you are really helpful.

Q.C : The angle about which the boom is inclined ( 60o in the example) doesn't make any difference for the tension in the support cable,so I am excluding it from the calculation? Then, F2= 200*9.81+250*9.81=4414.5 N ?

I don't understand what are they asking for in Q.A.. is it the radius of the ball?

 

[nvn]

Frisky90: Your answer for (c) is incorrect. The boom angle makes a difference in the support cable tension. Use method of sections. Cut the two cables, draw a free-body diagram (FBD) of the sectioned-off portion, and sum moments about point A. Point A is the pivot pin at the lower end of the crane boom (i.e., point A is the origin of the x and z axes). Solve the moment summation for F2.

Question (a) is asking for R2, the horizontal distance from the ball centerpoint to the z axis.

 

[Frisky90]

Frisky90: Your answer for (c) is incorrect. The boom angle makes a difference in the support cable tension. Cut the two cables, draw a free-body diagram (FBD) of the section-off portion, and sum moments about point A. Point A is the pivot pin at the lower end of the crane boom (i.e., point A is the origin of the x and z axes). Solve the moment summation for F2.

Question (a) is asking for R2, the horizontal distance from the ball centerpoint to the z axis.

Thx, this clarified the things a bit. So, the distance between the projection of the center of the ball on the horizontal and pivot would be 1/2*2.5=1.25 m which is the answer to Q.a.

Can you give me more info on how to sum the momments about A?

 

[nvn]

Frisky90: Your answer for question (a) is incorrect. You forgot about radius R. Try again. For question (c), use method of sections. You would need to draw a FBD, and show your calculations, so we will know what is causing you trouble. See big hints in post 8.

 

[Frisky90]

Frisky90: Your answer for question (a) is incorrect. You forgot about radius R. Try again. For question (c), use method of sections. You would need to draw a FBD, and show your calculations, so we will know what is causing you trouble. See big hints in post 8.

a) R2=1.25+0.2=1.45 ?

c) $$\sum$$ Fx=-4414.5+Fa(-->)=0
$$\sum$$ Fy=-4414.5cos60+Fr=0

 

[nvn]

Frisky90: (a) Your answer is correct. Nice work. (c) There is no FBD. And your current calculations appear to not make sense. You might want to study one or two example problems in your textbook, where they are performing a summation of moment. And study an example where they use method of sections. Also, you could upload a photograph of a quick, rough FBD to imageshack.us.

By the way, normally when you reply to a post, press the "New Reply" button at the top or bottom of the page, not the "Quote" button. You do not need to press the "Quote" button on every reply.

 

[Frisky90]

Ok,here is the FBD : http://img593.imageshack.us/i/img5989v.jpg/

Moment of the 2452.5 F = 1.45*2452.5=3556.125 N
Moment of the 1962 F= 0.625*1962=1226.25 N
Sum of moments= 4782.375 N = Ax (tension in support cable ? )

 

[nvn]

Frisky90: Note that FBDs usually have key points labeled with capital letters, and are usually dimensioned. Go ahead and label your key points, and dimension your FBD. Your answer for Ax is currently incorrect. (But call it T2, instead of Ax. Let T2 mean tensile force in the support cable. Call the tensile force in the lifting cable T1.)

T1 should not be attached directly to the tip of the boom. Draw the pulley on the boom upper tip; and move your T1 force to the correct location. Also draw and dimension the lifting cable tensile force for the lifting cable you cut, and include it in your moment summation.

You currently have the support cable attached to the wrong location. Move it to the correct location. And dimension your FBD. Finally, write a moment summation equilibrium equation about point A, and show it in your work. And then solve that equation for the support cable tensile force, T2.

 

[bunny1492]

how did u get the 1862 and .625 in the moment part?

 

[Frisky90]

New FBD: http://img200.imageshack.us/i/img5991sf.jpg/

M(T1)=2.452.2*1.45=3556.125 N
moment along the center of the boom=(1/2*2.5)*1/2(Right Triangle rule)*1962=1226.25 N
M(T2)=T2*1.5

3556.125+1226.25+T2*1.5=0
T2=-3188.25 ??

Pls tell me the exact way to do that if the answer is incorrect ,because this homework is due in 2 days.

 

[nvn]

Frisky90: Just this one time, I am showing you how to draw a FBD correctly. See file http://img51.imageshack.us/img51/6912/img5991sf01.png . Notice, you label key points with capital letters; and you show all dimensions. And you label all force vectors.

In post 16, you computed M(T1) and M(W1) correctly. Nice work. You computed M(T2) incorrectly. You need to determine distance z2. Also, you forgot to cut, draw, and dimension the lifting cable, as shown on the diagram -- and include it in your moment summation.

Now try your moment summation equilibrium equation again, summing moments about point A. And pay attention to the direction of each moment (positive or negative). Use the appropriate sign in your moment summation equation. Your sign on M(T2) is currently incorrect. Usually we define counterclockwise moment as positive, and clockwise moment as negative.

 

[Frisky90]

Thanks for the FBD.

I computed z2= 1.3m.
M(T1) along the lifting cable 0.2*T1= 490.5 N

4782.375-T2*1.3+490.5=0
T2=4056.06 N?

 

[nvn]

Frisky90: Notice, the second M(T1) is clockwise moment. Check the sign on this moment in your summation equation. Try again.

By the way, generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. Therefore, use z2 = 1.2990 m, not 1.3 m.

 

[Frisky90]

Ok, thanks for the advice.
I changed the sign and used 1.2990 for z2. So, the new answer is 3303.9838 ?

 

[nvn]

Your answer appears to be correct. To list a final answer (but not to reuse in calculations), you would generally round to three or four significant digits (3300 N, or 3304 N).

 

[Frisky90]

Thank you for the prompt responses.

In order to find the reactive pivot force (Q.d) I summ all the forces in the vertical

-T1-T(W)- T1cos60+T(R)=0
-2452.5-1962-1226.25+T(R)=0
T(R)=5640.75 N Correct?

 

[nvn]

Frisky90: Before you move on to (d), I just noticed a possible mistake we made. Therefore, save your current work, or a copy of it, in case it is correct. But I just noticed that the problem statement says the crane is rotating at angular velocity omega = 1.0 rad/s. This will make the demolition ball move outward, and will change the forces in the system. This makes the problem more complicated.

Is this a statics course? Or a dynamics course? You would need to know how to compute the centrifugal force due to circular motion, and the location of the ball, and the corresponding cable tensile force. In other words, you would need to have some knowledge of dynamics.

 

[Frisky90]

It is a bit of both. We are just starting dynamics.
I also noticed that it is said in the problem'' For the dynamic conditions described'' , so I guess this value should be used and the calculations you say should be made but I don't have any idea how.

 

[nvn]

You need to figure out the centrifugal force on the ball, and the position of the ball, and the cable tensile force T1. You will need to figure out the new R2, the new T1, and the new T2.

 

[Frisky90]

The centrifugal force on the ball is F=(mv^2)/r= 125 N
The tension in the cables will vary depending on the position of the ball. It will go back and forth 180 degrees parabola ??
I don't know how to connect this to the moments equation.