Simple Tension Problem: Calculating Tension and Acceleration

  • Thread starter Thread starter physicsnewb7
  • Start date Start date
  • Tags Tags
    Tension
AI Thread Summary
The discussion revolves around calculating the tension and acceleration of two masses (75 kg and 100 kg) suspended by a frictionless string on a frictionless pulley. The initial approach incorrectly assumed the acceleration of the heavier mass was a fraction of gravitational acceleration. Correctly applying Newton's second law reveals that the acceleration can be calculated as a = (25/175)g or g/7. The tension in the string can then be derived from this acceleration using the equations for each mass. Overall, the correct method involves setting up equations for both masses and solving them simultaneously.
physicsnewb7
Messages
42
Reaction score
0

Homework Statement


Two masses are suspended by a frictionless string on a frictionless pulley, one with mass 75 kg and the other 100 kg.
What is the tension of the string and the acceleration of the masses

The Attempt at a Solution


My attempt was to say since mass A, 75 kg, was 75 percent of mass B, then the acceleration of mass B would be 25 percent of the acceleration of gravity 9.8 m/s^2.
so it would be roughly 2.5 m/s^2. Since the acceleration was 2.5 m/s^2, I thought the tension would be equivelant to the force of the mass of both A and B times the acceleration 2.5 m/s^2 ie. F=ma. Clearly there is something wrong with this approach. What is it that I'm missing? is there a generalized formula for tension of a string?
 
Physics news on Phys.org
physicsnewb7 said:
Two masses are suspended by a frictionless string on a frictionless pulley, one with mass 75 kg and the other 100 kg.
What is the tension of the string and the acceleration of the masses

Hi physicsnewb7! :smile:

Call the tension T, and then apply good ol' Newton's second law twice (to each mass separately) …

since the acceleration, a, of each mass is the same (in opposite directions), you can solve for T and a :wink:
 
so if a I say a100kg=100g-T and a75=T-75g, where a is the acceleration, g is gravitational acceleration and T is tension, then add the two equations together to get a=25/175g=1/7g and then solve for T by plugging a in the equations. Is that right? would the acceleration be 1/7g? any help would be greatly appreciated!
 
physicsnewb7 said:
so if a I say a100kg=100g-T and a75=T-75g, where a is the acceleration, g is gravitational acceleration and T is tension, then add the two equations together to get a=25/175g=1/7g and then solve for T by plugging a in the equations. Is that right? would the acceleration be 1/7g? any help would be greatly appreciated!

Hi physicsnewb7! :smile:

Yes, that's right! :approve:

Another way of doing it is to imagine that the string is straight …

there's 100g of force pulling it left, and 75g pulling it right, making a total of 25g … since the total mass is 175, that makes a = 175g/25 = g/7. :wink:

(or, if you're not interested in the tension, you could even use conservation of energy and a = dv/dt = v dv/dh = 1/2 d(v2)/dh)
 
Thank you so much TinyTim for responding and helping.:smile:
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Tension in a string which connects 3 pulleys
Replies
12
Views
1K
Monkey accelerating up and down a rope
Replies
38
Views
4K
Problem with pulleys - two fixed and one movable
Replies
22
Views
371
Rocket + Connected Body Dynamics problem
Replies
47
Views
3K
Feynman's Lectures exercise about Newton's laws
Replies
13
Views
2K
Acceleration of rising mass and tension in cord
Replies
33
Views
3K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
2K
Finding tension of a string spinning horizontally
Replies
9
Views
2K
Tension force and acceleration Problem
Replies
1
Views
2K
How Do You Calculate the Tension of a Cello String?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top