What is the average pressure exerted on the other pedestal.

In summary, the sculpture is a small rectangular block made of granite with a weight of 102 kg. It will be balanced on two granite pedestals, one with a contact area of 25 square centimeters and the other with a contact area of 13 square centimeters. The first pedestal sustains an average pressure of 2kg/square centimeter, therefore the other pedestal must support the remaining weight of 52 kg, resulting in an average pressure of 4kg/square centimeter.
  • #1
lokal704
10
0

Homework Statement


An artist has designed a sculpture made from granite. The sculpture is essentially a small rectangular block whose weight is 102 kg. This centerpiece will be balanced on two granite pedestals, one at each end of the centerpiece. The pedestals are different sizes, such that one pedestal has a total area of contact with the centerpiece of 25 square centimeters and the other has an area of contact of 13 square centimeters. The first pedestal (area = 25 square centimeters) sustains an average pressure of 2kg/square centimeter. (Here "kg" means the force exerted by a mass of 1 kg.) What is the average pressure exerted on the other pedestal. Show Your Work!

Any help would be great. Thanks!



Homework Equations





The Attempt at a Solution

 
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  • #2


lokal704 said:
Show Your Work!

Any help would be great. Thanks!

Actually any work shown would be great.

Thanks.
 
  • #3


Oh, sorry. This is the second time I have used this forum, and I am still new.

Here is what I have so far:

p=F/A

First answer:

2=F/25
2*25=F
F=50

so would it be

p=50/13
p=3.85 kg?
 
  • #4


lokal704 said:
Oh, sorry. This is the second time I have used this forum, and I am still new.

Here is what I have so far:

p=F/A

First answer:

2=F/25
2*25=F
F=50

so would it be

p=50/13
p=3.85 kg?

Almost. According to the statement you have 2 lbs over 25 cm2 and as you point out that represents 50 lbs overall.

But doesn't the object actually weigh 102 kg and hence doesn't the other pedestal need to provide for all of the remainder?
 
  • #5


SO would it be the remaining 52 kg? Or 102 kg?
 
  • #6


lokal704 said:
SO would it be the remaining 52 kg? Or 102 kg?

Well let's think about it.

If one end is holding up 50 kg, what will the other end hold up? What must the two ends support totally if it is at rest?
 
  • #7


it would be 52 right? Because collectively they are supporting 102. So it should look like this right? Please correct me if I am wrong.

p=52/13

p=4kg
 

Related to What is the average pressure exerted on the other pedestal.

Question 1: What is the definition of "average pressure"?

The average pressure is the force applied per unit area of a surface. It is calculated by dividing the force exerted on an object by the area over which the force is distributed.

Question 2: How is the average pressure exerted on the other pedestal calculated?

The average pressure exerted on the other pedestal is calculated by dividing the total force exerted on the pedestal by the area of the pedestal's surface in contact with the object. This will give the pressure exerted on that specific area.

Question 3: Is the average pressure exerted on the other pedestal the same as the pressure exerted on the entire surface?

No, the average pressure exerted on the other pedestal is only calculated for the specific area in contact with the object. The pressure exerted on the entire surface may vary depending on the size and shape of the object and the distribution of the force.

Question 4: What units are used to measure average pressure?

The SI unit for pressure is the Pascal (Pa), which is equivalent to one Newton per square meter (N/m²). Other common units for pressure include pounds per square inch (psi) and atmospheres (atm).

Question 5: Can the average pressure exerted on the other pedestal change?

Yes, the average pressure exerted on the other pedestal can change depending on the amount of force applied or the area in contact with the object. It can also change if the object or pedestal moves or if the distribution of the force changes.

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