Tension headache!

  • Thread starter brenfox
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  • #1
brenfox
71
1

Homework Statement


a mass of 0.5kg is suspended from a flywheel. if the mass is released from rest and falls a distance of 0.5m in 1.5s, calculate : The tension in the rope. acceleration is 0.444ms-2.


Homework Equations


t-w = -ma


The Attempt at a Solution



t = m(ag)

t = 0.5(0.4444=9.81)

t = 5.127n

w = mg = 0.5x9.81 4.9n

so t-w 0.227n is rope tension.

Check by inserting numbers into equation of t-w = -ma

so: 5.127 - 4.905 = 0.5 x 0.222.

The last equation doesn`t marry up?
 

Answers and Replies

  • #2
janhaa
97
3
The tension in the rope is:

T = m(g - a) = 0,5(9,81 - 0,44)

or...?
 
  • #3
abrewmaster
36
5
If you had the mass just sitting there with no acceleration then it would be simply T=mg
Since it's accelerating you have to subtract the force acting on it from the acceleration that it is going at which is F=ma so therefore:
T=mg-ma=m(g-a)=0.5(9.81-0.444)=4.683N should be your answer?
 
  • #4
brenfox
71
1
From the beginning

t-w = -ma

t=m(g-a) = 0.5(9.81-0.44)

= 4.683n so t = 4.683n

w=mg so w = 0.5x9.81 = 4.905n

S0 t-w = -ma

Checking...... 4.683-4.905 = -0.5 x 0.444

Correct... i think.
 
  • #5
CAF123
Gold Member
2,950
88

Homework Statement


a mass of 0.5kg is suspended from a flywheel. if the mass is released from rest and falls a distance of 0.5m in 1.5s, calculate : The tension in the rope. acceleration is 0.444ms-2.


Homework Equations


t-w = -ma


The Attempt at a Solution



t = m(ag)

t = 0.5(0.4444=9.81)

t = 5.127n

w = mg = 0.5x9.81 4.9n

so t-w 0.227n is rope tension.

Check by inserting numbers into equation of t-w = -ma

so: 5.127 - 4.905 = 0.5 x 0.222.

The last equation doesn`t marry up?

It is convenient to denote the positive direction in the direction of acceleration. In this case, the mass accelerates downwards so take down to be positive. This gives mg - T = ma. Solving gives T = m(g-a). It is difficult to see where you went wrong in your attempt.
 

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