Newton's Third Law of toboggans

[Dough]

Three toboggans are attached to each other with ropes. A force of 145 N[forward] is pulling the first toboggan, which has a mass of 42kg. A second toboggan, with a mass of 30kg, is attached to the first second toboggan. A third toboggan with a mass of 24kg, is attached to the second toboggan. Assume that the surface is frictionless.
a) What is the acceleration of all three toboggans?
b) Calculate the tension in the rope between the first toboggan and the second toboggan.
c) Calculate the tension in the rope between the second toboggan and the third toboggan.

i would like to verify my answers and if you guys could show hwo you do b and c step by step or explain it, it would help
B) 81.54N
C) 36.24N


Please and thank you :D

 

[James R]

Your answers are correct.

It is all a matter of how you define the "system" that you're looking at. For any system, we have:

F(net external) = M(total) a(centre of mass)

For part (a), you can define the system as including all three toboggans. The tension forces are internal forces, and the only external force is the 145 N force. The total mass is the sum of the three masses of the toboggans making up the system.

For part (b), for example, take instead just the first toboggan as the system. There are two external forces on this system: the 145 N force and the tension force from the connection to the 2nd toboggan. So, for this system:

145 - T = 42a

You already know a from part (a), so you can solve for T.

It is also possible, for example, to consider a system consisting of two of the three toboggans.

Hope that helps.

 

[Dough]

yes it does, i liek to think of things logically (it makes it easier to remember what forumla to use and stuff) so, the tension in the connection between the first and 2nd toboggan would be higher than the last and 2nd one because there is more weight being pulled... correct?