Ask a question

Physics Another Physics question

Three objects are connected by light strings as shown in Figure P4.62. The string connecting the m1 = 5.50 kg mass and the m2 = 4.20 kg mass passes over a light frictionless pulley.

Figure P4.62 Image of figure in link (  )

(a) Determine the acceleration of each mass.
            m/s2 (magnitude)
(b) Determine the tension in the two strings.
            N (string between m1 and m2)
            N (string between m2 and the 3.00 kg mass)

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Sung taee L. | Teach Concepts, Thinking methods, Step by Step (Physics & Math)Teach Concepts, Thinking methods, Step b...
5.0 5.0 (20 lesson ratings) (20)

If you draw free body diagrams for each mass, you will obtain these three equations. (Sorry I can not show you the diagrams) For each diagram, you have to apply the Newton's 2nd law.

1) for m3: 3g - T3 =3a (T3 is the tension between m2 and m3, "g" is gravitational acceleration = 9.8 m/s and "a" is the acceleration of the three masses which falling to the m2 and m3 side)

2) for m2: 4.2g + T3 -T2 = 4.2g (T2 is the tension between m1 and m2)

3) for m1: T2 - 5.5g = 5.5a (Please be careful when you write this eaquation, you have to consider the direction of acceleration. Put the direction of acceleration as a positive direction)

If you substitute 9.8 for "g" and solve the 3 linear first order equations then you can obtain the answers.

(a) acceleration a = 1.312 m/s2 for each mass

(b) T2 = 61.116 N and T3 = 25.466 N

The problem wants up to 2nd decimal position so you have to round the given answers to fit the given significant numbers. If there are any required conditions, you have to fit that requirement related with the significant numbers.

I hope you can understand this question without diagrams. I feel sorry about the limitation of this answering box.


Kerrie T. | math and science from a professional academic scientistmath and science from a professional aca...
4.8 4.8 (332 lesson ratings) (332)

I deleted this answer and tried to post a corrected comment with better formatting...and a sign correction (thanks sung!)


Wow, I had a problem with the answer box, here, too... let me try to reorganize a bit...

You have to draw a force diagram to answer this question.

    • First, use your intuition to set things up: You know that the masses are moving, and that they are moving such that the heaviest weight (m1) is going down. right?
  • Draw a force diagram for m1. The weight pulls down and the string pulls up. T1 (up) - m1*g (down) =  m1*a1 (up) (cheesy diagram follows)





                                  O   m1





  • Any acceleration is caused by unbalanced forces. So a1 is the answer to part (a), but we don't have enough information to solve yet, until we figure out the value for T1.
    • More common sense: Two moving objects attached by a taught (under tension) string must be moving at the same speed and acceleration, otherwise the rope would be slack.  So call the acceleration a instead of a1.
    • AND The tension in the string is uniform throughout, so T1 must be equal to the tension on the other end, so I'm just going to call that T1, too. So the forces acting on m2 are a little more complicated, but along the same lines:
  • T1 (up) - T2 (down) - m2*g (down) = - m2 * a (down) ....
    • remember m1 and m2 are experiencing the same acceleration ...but m2 is going down. Now we have another equation AND another variable, so lets keep going...
  • The last object is the 3kg bucket. T2 (up) - 3kg * g (down) = -3kg * a (up)
    • AHA, finally an unknown we can actually solve for.Rearrange that last equation.
  • T2 = 3kg *g - 3kg *a
    • now substitute that into the second equation:
  • T1 - T2 - m2*g = -m2*a will become
  • T1 -{3*g -3*a} -m2*g= -m2*a, and if we simplify and rearrange,
  • T1 = 3*g -3*a +m2*g - m2*a Now substitute that into the first equation:
  • T1 - m1*g = m1*a becomes
  • {3*g +m2*g -3*a - m2*a} -m1*g = m1*a
    • Now we need to group all the a terms, and put others on the opposite side of the equal sign.
  • 3*a +m2*a +m1*a = 3*g +m2*g -m1*g
  • a*(3+m2+m1) = (m1 -3 -m2)*g divide both sides to solve for a
  • a={(m1 - m2 - 3)*g} / {3+m2+m1} You can plug in values to solve
  • a= (5.5-4.2-3)*9.81 / (3+5.5+4.2) = (-1.7 *9.8) / 12.7 = 1.31
    • The units for a are in meters per sec squared, because those were the units for g=9.8
  • Now you can solve for part (b)
    • T1 - 5.5*9.81 =  5.5*1.311 can be rearranged
  • T1= 5.5*9.81 +5.5*1.31 = 5.5 *11.1= 61.2 Newtons
    • AND you can solve for  T2 =T2 = 3 *g - 3 *a
  • = 3 *(g - a) = 3*(9.81 - 1.31) = 25.5 Newtons.