Calculating the mass of an object pivoting on a point

In summary, the pivot point on a 2.2m uniform beam is 0.50m from one end (labeled A). The beam can be balanced on the pivot by placing an 8.0kg load at end A. To determine the mass of the beam, it is important to realize that the clockwise torque must be equal to the counterclockwise torque for the beam to balance. The counterclockwise torque is equal to the force of the 8kg load multiplied by the distance from the pivot (0.5m), which results in a torque of 39.2N*m. The clockwise torque is due to the weight of the beam, which can be considered as acting through the center of gravity of
  • #1
agm1984
25
0

Homework Statement



A pivot is 0.50m from the end (called A) of a 2.2m uniform beam. A load of 8.0kg at end 'A' will cause the beam to be balanced on the pivot. What is the mass of the beam?

A-----------------------
...Δ

Homework Equations



τ = d*f

The Attempt at a Solution



im close to understanding this but i just don't see how they deduce the mass of the beam.

from what i understand, there are a couple critical things to realize:
1) the clockwise torque = counterclockwise torque because its balanced (which means the torque on both sides are equal)

8kg*9.8=78.4N (this is the counterclockwise force 0.5m away from pivot, thus producing 39.2N*m torque

so the clockwise torque will be equivalent. the end of the beam to the right is 1.7m from the pivot and i don't know exactly how relevant that is.

2)additionally, i later realized that the center of gravity will be at 1.1m (halfway due to uniform beam) and I am not sure but i suspect there is 9.8N pushing counterclockwise here.

i just don't see how i take all this information and deduce the mass of the beam. i would really appreciate some help, preferably a step by step solution using words rather than numbers.
 
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  • #2
OK
Well you are on the right track.
On one end labelled A of the beam there is the 8kg mass and the mass of the beam from pivot P to A, of length 0.50m. Both the 8.0kg and mass of beam PA provide a torque.
From the pivot P to the other end, label it B, there is the mass of the beam of length 1.70m, providing a torque.
Torques on both sides of the bean must be equal for the beam to balance.
Does that help?
 
  • #3
Hello agm1984. Welcome to PF!

You've got the counterclockwise torque correct.

The key point (that you are close to getting) is that the entire weight of the beam can be considered as acting through the center of gravity of the beam for the purpose of calculating the torque due to the weight of the beam. So, with that in mind, see if you can determine what the weight of the beam must be to provide the clockwise torque that will balance the counterclockwise torque.
 
  • #4
thanks for your reply. I am not sure if i understand now or not...

from your reply i set up the equation (first: 8kg*9.8N=78.4N)

removing all the units for clarity sake

this below is essentially Fd + Md = Md (where F is the 8kg, and M is the mass of the beam, and 'd' is distance)
(78.4)(0.5) + M(0.5) = M(1.7)
when i solve for M i get 32.666 repeating Newtons, and dividing by 9.8 i get 3.333333 repeating kilograms

but i still don't understand where I am at. but i do notice that we have 3.3kg on both sides. do i just add the two sides together 3.33 + 3.33 to get 6.66 (rounded up to 6.7kg which is the answer in the book?)

if so is that how you set up the question for a solution?
 
  • #5
agm1984 said:
(78.4)(0.5) + M(0.5) = M(1.7)
Two things wrong with that. You're treating the parts of the beam each side of the pivot as equal in mass to the whole beam; and you're treating those masses as concentrated at the ends of the beam.
There is a simpler way. You can treat the whole beam in one go as acting at its mass centre. Logically splitting it into separate masses divided at the pivot will produce the same answer, if done correctly.
 
  • #6
haruspex said:
You can treat the whole beam in one go as acting at its mass centre. Logically splitting it into separate masses divided at the pivot will produce the same answer, if done correctly.

i don't understand what you are saying

oh ok wait maybe i do

so there the force of 8kg 0.5m... and since the center of gravity is at 1.1m can we say that all of the balancing force is at 1.1m away from A? (which is 0.6m to the right of the pivot point)

so the equation would be like...
-> Fd = Fd
-> (78.4N)(0.5m) = (FN)(0.6m)
-> 39.2 = 0.6F
-> F=65.333repeating Newtons

-> F/9.8=kilograms

kilograms=6.66repeating (which is the answer when taken to two sig figs)

AHA i get it now... that is to say i might... if i interpretted that correctly and my answer isn't a cooncidence
 
Last edited:
  • #7
i'd appreciate if someone could confirm that my last attempt is the logically correct way. i just bolded the portion of my last post that i am most concerned with.
 
  • #8
agm1984 said:
i'd appreciate if someone could confirm that my last attempt is the logically correct way. i just bolded the portion of my last post that i am most concerned with.

Yes, it's correct. Good.
 

Related to Calculating the mass of an object pivoting on a point

What is the formula for calculating the mass of an object pivoting on a point?

The formula is: mass = (distance to pivot point) x (force applied) / (acceleration due to gravity).

Can you explain the concept of torque in relation to calculating the mass of a pivoting object?

Torque is the measure of the turning force on an object. In order to calculate the mass of a pivoting object, we need to consider the torque applied at the pivot point, which is equal to the force applied multiplied by the distance from the pivot point to the point where the force is applied.

What is the role of acceleration due to gravity in calculating the mass of a pivoting object?

Acceleration due to gravity, denoted as "g", is a constant value that represents the rate at which objects accelerate towards the Earth. In the formula for calculating the mass of a pivoting object, g is used to convert the force applied into a torque value.

What units should be used when calculating the mass of a pivoting object?

The units used will depend on the units used for distance (usually meters) and force (usually newtons). The resulting unit for mass will be kilograms.

Are there any common mistakes to avoid when calculating the mass of a pivoting object?

One common mistake is forgetting to convert the force applied into a torque value by multiplying it by the distance to the pivot point. It is also important to use consistent units throughout the calculation and to ensure that the acceleration due to gravity value used is appropriate for the location where the calculation is being performed.

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