help with physics problem

[Jonesy]

Ok this was given as a practice problem for my test next week and also extra credit. I've tried working on it for about an hour but can't seem to get it to work out. My teacher can't teach so, i need some help.

I'll try to scan or use paint to post a free body diagram that was given tomorrow.

1) Calculate how much force the jaw of a robotic gripper must apply to hold a part in a vertical plane under the following conditions.
a) Part weighs 60 pounds and has a non uniform shape
b) The gripper jaws are parallel to each other
c) the part is grasped 20 inches from it's center of gravity
d) the jaws gripping the surface are 15 inches wide
e) the part is 1.75 inches thick at the point where it is being gripped.
f) the part is being lifted with a maximum acceleration of 4 Gs including normal gravitational force
g) a coefficient of friction between the part and gripper of 0.65
h) a safety factor of 2 must be included.

2) what would the required force be if the gripper operates under the same conditions as in question 1, but in a horizontal plane.


I keep getting really huge numbers that don't make any sense.


Thanks for any help, or anything to point me in the right direction

[honduh440]

7 feet

[Aceman]

WILKIN!! Where you at??:D

[44oEX]

well the way I see the probleme I get F=1028.92N

but I could be wrong

for #1

for # 2 I think you would need to know the shape

[derekhonda]

My roomate is a physics major, but so far all he has learned is that a cat has to fall from 6 stories or higher to reach terminal speed. Damn handy information though.

[zephead400ex]

Originally posted by derekhonda
My roomate is a physics major, but so far all he has learned is that a cat has to fall from 6 stories or higher to reach terminal speed. Damn handy information though.

LMFAO!

[wilkin250r]

Originally posted by Aceman
WILKIN!! Where you at??:D

I would need to see the diagram.

It would seem to me, that if you are holding the part in a verticle plane, the only forces acting on it are the friction of the jaws, and gravity. Do you have a torque associated with it, to keep it from swaying? That's the ONLY reason I can think for needing the width of the jaws, and the location of the grip.

Is this a two-dimensional problem, or three?

[Quad18star]

Originally posted by wilkin250r
I would need to see the diagram.

It would seem to me, that if you are holding the part in a verticle plane, the only forces acting on it are the friction of the jaws, and gravity. Do you have a torque associated with it, to keep it from swaying? That's the ONLY reason I can think for needing the width of the jaws, and the location of the grip.

Is this a two-dimensional problem, or three?

^^^ What he said !!!!

2 years of physics and physics related classes , and I still need to see diagrams . Are you sure you gave us all the information also ??

[wilkin250r]

I'm assuming the jaw is not on a cable, and does not pivot.

I'm also assuming that the center of gravity is NOT directly inline with the jaw, and thus you have a torque associated with it, and you need to keep the part from swaying or pivoting. Otherwise, you don't need the width of the jaws, or the irregular shape of the part.

As such, if the center of gravity is not directly inline with the jaw, I need more information than "20 inches". I need X and Y coordinates, or an angle.

I have NO idea why you would need the thickness of the part where it is gripped.

[Quad18star]

Originally posted by wilkin250r
I'm assuming the jaw is not on a cable, and does not pivot.

I'm also assuming that the center of gravity is NOT directly inline with the jaw, and thus you have a torque associated with it, and you need to keep the part from swaying or pivoting. Otherwise, you don't need the width of the jaws, or the irregular shape of the part.

As such, if the center of gravity is not directly inline with the jaw, I need more information than "20 inches". I need X and Y coordinates, or an angle.

I have NO idea why you would need the thickness of the part where it is gripped.

My college professors used to give us useless information , just to screw us up , and make you think if you need it or not . Maybe he gave the thickness because of sheering ??? I dunno !!!! I need to see a picture to be able to figure this one out ... I hate word problems .. it's my weakness !!!

[tore up]

Originally posted by derekhonda
My roomate is a physics major, but so far all he has learned is that a cat has to fall from 6 stories or higher to reach terminal speed. Damn handy information though.

your right that is damn handy info!!!!!!!:macho

[44oEX]

well wilkins the way I see it is that the #1 is just a jaw.that is mounted on a solid surface. that is holing the piece that is poiting staight down. so there if

-masse X gravity = -static friction + acceleration

and for #2 the center of gravity that is 20 inch from the end of the jaw. so the 60 pounds X 20 inch is causing a torque on the jaw. you would need the lenght of the jaw cause the weight would try to move the lower piece down and the force would be at the end of the jaw and the force on the upper part would be at the start of the jaw ( toward the mounting area. )

[Jonesy]

These problems are for my robotics class. We are trying to find out the amount of force (in pounds) for each jaw that is required to hold the object in the respective plane, then multiply it by 2 for the safety factor.

I have the drawings and will scan and post them once i get off work. He also gave us several equations that i will post.

Thanks for the help.

Jonesy

[pro_stock_1000]

dont listen to 44o ex i no the guys and is only in is firt year in nursing.......so hes probably wrong.....

[44oEX]

Originally posted by pro_stock_1000
dont listen to 44o ex i no the guys and is only in is firt year in nursing.......so hes probably wrong.....

be nice know. I am in 2nd year in mechanical eng.