Question

The rope will snap if the tension in it exceeds 55.0 n. what is the maximum value of the force f which can be applied?

Answer 1

Missing part of the text:
"Two masses, m1 = 2.12 kg and m2 = 9.01 kg are on a horizontal frictionless surface and they are connected together with a rope as shown in the figure."
and missing figure (see attachment)

Solution:
We can write Newton's second law for the whole system m1-m2 and for m2 only (2 equations). Only one force (F) acts on the m1-m2 system, while if we consider m2 only we have two forces acting on it: F and T (tension), in the opposite direction. So, the two equations are

`F=(m_1+m_2)a`

`m_2 a=F-T`
where a is the acceleration of the system.

From the first equation we get

`a= (F)/(m_1+m_2)`
and substituting it inside the second equation, we get

`(m_2)/(m_1+m_2)F=F-T`
re-arranging, we get

`F=T( (m_1+m_2)/(m_1) )`

Using
`m_1=2.12 Kg`,
`m_2=9.01 Kg`, and using the maximum value of T that is allowed not to break the rope (T=55 N), we can find the maximum allowed value for F:

`F=288.8 N`