How Do You Calculate the Tension in a String Lifting an Accelerating Mass?

  • Thread starter Thread starter iamtrojan3
  • Start date Start date
  • Tags Tags
    Lifting Mass
AI Thread Summary
To calculate the tension in a string lifting an accelerating mass, first determine the weight of the mass using the formula f = ma, resulting in 16.17N for a 1.65 kg mass under gravity. The upward acceleration of 3.10 m/s^2 adds an additional force of 11.05N. The correct approach involves recognizing that the tension must counteract both the weight and provide the force for the upward acceleration. By applying the free body diagram concept, the relationship between tension, weight, and resultant force can be established, leading to the correct tension calculation. Understanding these forces is crucial for solving the problem accurately.
iamtrojan3
Messages
55
Reaction score
0

Homework Statement


A 1.65 kg mass is suspended from a string which is pulled upward. The mass accelerates upwards with an acceleration of 3.10 m/s^2. What is the tension in the string?

Homework Equations



f = ma

The Attempt at a Solution


seems really easy but i can't get the right answer

the regular tension would be 1.65kg * 9.8m/s^2 = 16.17N

if its pulled up 3.1 m/s^2, the forces in that would be 1.65kg * 3.1m/s^2 = 11.05N

then i subtracted the forces to get 5.1, but its wrong.

that wasn't right so i added the forces, still wrong.

then i did the acceleration as a whole, as in 9.8 + 3.1 = 12.9
then found the forces which is 21.3N... still wrong

anyideas? thanks
 
Physics news on Phys.org
Draw a free body diagram, and solve the problem as you see the forces there (gravity, upward acceleration included). The acceleration would be modified.
 
iamtrojan3 said:
seems really easy but i can't get the right answer

the regular tension would be 1.65kg * 9.8m/s^2 = 16.17N

yes regular if it was in equilibrium, so just know that is the weight of the mass.


iamtrojan3 said:
if its pulled up 3.1 m/s^2, the forces in that would be 1.65kg * 3.1m/s^2 = 11.05N

3.1 m/s2 is the resultant acceleration, multiplying it by the 1.65kg gives the resultant force.



In your free body diagram, the tension acts upwards while the weight acts downwards. Can you make a relationship between the tension,weight and resultant force? If you can do that, then the question become quite easy.
 
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}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...

Similar threads

Replies
38
Views
4K
Replies
33
Views
3K
Replies
13
Views
3K
Replies
12
Views
1K
Replies
8
Views
2K
Replies
4
Views
4K
Replies
1
Views
2K
Back
Top