Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page Visit Our Archives Ask A Question How To Ask A Question Question of the Week NEWTON Teachers Our Expert Scientists Volunteer at NEWTON! Referencing NEWTON Frequently Asked Questions About Ask A Scientist About NEWTON Education At Argonne Truncated Cone and Support

Name: Nicole
Status: educator
Grade: 9-12
Location: VT
Country: USA
Date: Fall 2013


Question:
Below is a question that one of my students just ask. I feel like it should matter if the cup is upside down or right side up because if it is sitting on a table it will have forces from the object sitting on it and from the table below. If we do not consider the weight of the cup, the force down from the object on the cup should be same as the force up from the table so the cup is being crushed from both sides. Here is the question: Imagine a plastic cup that has a diameter that increases as the height increases. The mouth of the cup is wider than the bottom. Could this cup support more weight upside down or right side up?



Replies:
Hi Nicole,

In theory it should not matter. Whether the cup is upside down or right side up, there is a force balance. If you put a 20 N load on the cup, that 20 N load is carried all the way through the cup from the top to the bottom. That same 20 N is felt by both the top of the cup and the bottom no matter the orientation of the cup.

Regards, John C Strong


Hi Niclole,

The cup (as you indicate) will have identical forces at each end. So, considering this fact, it is obvious that whether the cup is upright or upside down, each end will "see" no difference in the forces acting on it, no matter which way it is oriented. Therefore, there can be no difference in the cup's ability to support weight, no matter which way it is oriented.

Regards, Bob Wilson


Nicole,

If we ignore the weight of the cup, then the orientation up or down should not matter. If the cup is not moving, then all the forces must balance. The weight creates a force F=mg where m=mass of weight and g=gravitation field intensity (about -10 N/kg). Consider the two sides of the cup (in either orientation):

1. Side touching weight. The weight is still exerting a force F down whether it is on the table or on the cup. The cup must push back up with the same force; otherwise the weight would fall. 2. Side touching the table. If you place the cup and the weight on a scale (as a substitute for the table), it will not matter which direction the cup is oriented. It will measure the same weight, or force, on it. Thus, the table must also push back with the same force as the weight pushes down; otherwise again, the cup would move.

Thus, the two sides will feel the same force, regardless of cup orientation. Now, this force is not distributed equally on one side or the other. The question then becomes, "what side will collapse first?"

Kyle J. Bunch, PhD, PE



Click here to return to the Engineering Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 223
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: November 2011
Weclome To Newton

Argonne National Laboratory