Masses hanging on massless meter stick problem

In summary, for part b, in order to calculate the size of the angular acceleration of the meter stick at that instant, you need to use the formula angular acceleration = torque/inertia. In this case, the inertia is equal to the mass of the meter stick (3 kg) multiplied by the distance from the pivot (0.75 m) squared. The torque can be calculated by finding the net torque (which is equal to zero) and solving for the unknown force. Once you have both the torque and inertia, you can plug them into the formula to find the angular acceleration. The correct answer for this problem is 8.711 rad/s^2.
  • #1
FlipStyle1308
267
0
Two masses (mA = 3 kg, mB = 6 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.

a. Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level?

b. Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant.

For a I was clueless on how to approach the problem, but for b, I used angular acceleration = torque/inertia, and got 0.8889, but got it wrong. I tried it again, plugging in different values for r and F, and got 13.066, and still got the problem wrong. I thnk my problem is that I am not using the correct values to solve this problem. Anybody know exactly which values I should use? Thanks!
 
Physics news on Phys.org
  • #2
FlipStyle1308 said:
Two masses (mA = 3 kg, mB = 6 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.

a. Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level?

b. Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant.

For a I was clueless on how to approach the problem, but for b, I used angular acceleration = torque/inertia, and got 0.8889, but got it wrong. I tried it again, plugging in different values for r and F, and got 13.066, and still got the problem wrong. I thnk my problem is that I am not using the correct values to solve this problem. Anybody know exactly which values I should use? Thanks!

as it stands, the question does not make any sense. there is information missing. The ruler is resting on a pivot somehwere? Is it supported at the center??
 
  • #3
Sorry...I took out the parts I solved already...maybe this will help...

The system is then hung from a string, so that it stays horizontal. The string is placed 50 cm, from mass A.
 
  • #4
FlipStyle1308 said:
Sorry...I took out the parts I solved already...maybe this will help...

The system is then hung from a string, so that it stays horizontal. The string is placed 50 cm, from mass A.

Ok. Without that information it was impossible to answer.

You must impose that the net torque is zero. What is the torque exerted by mass A? The other torque must be minus that. Knowing the force, it is easy to find how far it must be from where the string is attached
 
  • #5
Wait, you confused me...which part are you on, a or b?
 
  • #6
FlipStyle1308 said:
Wait, you confused me...which part are you on, a or b?

well, I was doing part a first :approve:
 
  • #7
Okay, so part a haha. Torque exerted by mass A is 22.05, right?
 
  • #8
FlipStyle1308 said:
Okay, so part a haha. Torque exerted by mass A is 22.05, right?

?? How did you get this? torque = force * distance from the fulcrum and force = mg
 
  • #9
distance = 0.75
mass = 3
gravity = 9.8

I probably have the distance wrong, should it be 0.5?
 
  • #10
FlipStyle1308 said:
distance = 0.75
mass = 3
gravity = 9.8

I probably have the distance wrong, should it be 0.5?
Yes. torque = F d (in magnitude..the sign tells you the direction) The distance there is always the distance between where the force si applied and the axis of rotation (if the force is perpendicular to the line connecting the fulcrum and the point where the force is applied)
 
  • #11
Okay, so 14.7 N for torque.
 
  • #12
FlipStyle1308 said:
Okay, so 14.7 N for torque.

ok. The sign depends on whether th emass is to the left or right of the suspension point. Let's say it is to the left and creates a counterclockwise torque, then it is positive.

The other force will create a torque of - F * 0.50 m (since it is also 50 cm from the axis of rotation) Imposing that the sum of the two gives a net torque euqal to zero gives you the value of F (in that case, it is trivial because the two forces act at the same distance from the axis of rotation! so of course you will find a value equal to mg !)
 
  • #13
So I'm correct?
 
  • #14
FlipStyle1308 said:
So I'm correct?
yes, your torque is correct
 
  • #15
Thank you! So what do I do from there lol?
 
Last edited:
  • #16
FlipStyle1308 said:
Thank you! So what do I do from there lol?

My post #12 gives the way to answer part a.

I have to log off soon. Good luck
 
  • #17
Oh okay, I got part a, let me know when you can work on part b!
 
  • #18
Hmm...okay, for this one I tried...

I = mr^2 = (3)(0.75)^2 = 1.6875
angular acceleration = torque/I = 1.47/1.6875 = 8.711 rad/s^2.

Is this correct?
 

1. What is a "masses hanging on massless meter stick problem?"

The "masses hanging on massless meter stick problem" is a physics problem that involves calculating the forces and torques acting on a meter stick that is balanced on a pivot point and has two masses hanging from it.

2. What are the key concepts involved in this problem?

The key concepts involved in this problem are equilibrium, torque, and center of mass. Equilibrium refers to a state where the forces and torques acting on an object are balanced, resulting in no net movement. Torque is a measure of the rotational force acting on an object, while center of mass is the point where the mass of an object is evenly distributed.

3. How do you approach solving this problem?

To solve this problem, you would first draw a free body diagram to identify all the forces acting on the meter stick. Then, you would use the equations for equilibrium and torque to set up a system of equations and solve for the unknown variables. Finally, you would check your answer by ensuring that the forces and torques are balanced and the meter stick is in equilibrium.

4. What are some common mistakes made when solving this problem?

One common mistake is forgetting to consider the weight of the meter stick itself, which can affect the equilibrium and torque calculations. Another mistake is using the wrong units or not converting units consistently throughout the problem. It is also important to double check the direction of forces and torques to ensure they are correctly accounted for in the equations.

5. How is this problem relevant in real-world applications?

This problem is relevant in many real-world applications, such as designing bridges and buildings to withstand various loads, analyzing the stability of structures like cranes or scaffolding, and understanding the physics behind simple machines like seesaws and levers. It also helps develop critical thinking and problem-solving skills that can be applied to other areas of science and engineering.

Similar threads

Finding mass of unbalanced meter stick from torque
    • Introductory Physics Homework Help
Replies
2
Views
1K
Hanging mass on a massless pulley
    • Introductory Physics Homework Help
Replies
2
Views
884
Moment of inertia of a meter stick
    • Introductory Physics Homework Help
Replies
9
Views
5K
Finding the Mass of a Meter Stick (Torque and Balance)
    • Introductory Physics Homework Help
Replies
13
Views
6K
Calculate Weight of Block & Meter Stick | Torque Problem Solution
    • Introductory Physics Homework Help
Replies
1
Views
3K
Motion of center of mass under gravity
    • Introductory Physics Homework Help
Replies
10
Views
1K
Meter stick and rock balancing ?
    • Introductory Physics Homework Help
Replies
2
Views
4K
Finding mass of a meter stick using torque
    • Introductory Physics Homework Help
Replies
33
Views
20K
Fluids: Concept about Fluid Pressure in a relation with Force and Area
    • Introductory Physics Homework Help
Replies
6
Views
1K
Finding the Center of Mass of a Meter Stick with Multiple Masses
    • Introductory Physics Homework Help
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top