Confusion on Body Definition in Y&F, 9th, 4-49.

[Ackbach]

Homework Statement

From Young and Freedman's University Physics, 9th Ed., Problem 4-49. Two blocks are connected by a heavy uniform rope with a mass of $4.00$ kg. An upward force of $200$ N is applied to the top block, which has a mass of $6.00$ kg. The lower block has a mass of $5.00$ kg. a) Draw a free-body diagram for the $6.00$-kg block, a free-body diagram for the $4.00$-kg rope, and a free-body diagram for the $5.00$-kg block. For each force, indicate what body exerts that force. b) What is the acceleration of the system? c) What is the tension at the top of the heavy rope? d) What is the tension at the midpoint of the rope?

Homework Equations

$\vec{F}=m \vec{a}.$

The Attempt at a Solution

a) The $6.00$-kg block has three forces: the $200$-N force up, $mg$ down, and the tension in the rope pulling down. The rope also has three forces on it: the $200$-N force up, $mg$ down, and the weight of the lower block pulling down. The lower block has the tension force up, and $mg$ down.

b) The acceleration of the system we may compute by
$$a=\frac{200\, \text{N}-(15.00\, \text{kg})(9.8\, \text{m/s}^{2})}{15.00\, \text{kg}}=3.53
\, \text{m/s}^{2}.$$
This is in the upward direction.

c) To find the tension in the rope at the top, we must sum the forces and
use Newton's Second Law. That is, we know that
$$F-mg-m_{5}g=(m+m_{5})a,$$
or
$$F=(m+m_{5})a+mg+m_{5}g=(9.00)(3.53)+(4.00)(9.8)+(5.00)(9.8)=120\, \text{N}.$$

d) We compute the tension at the bottom of the block by using
$$F-m_{5}g=m_{5}a \implies F=m_{5}(a+g)=5(3.53+9.8)=66.65 \, \text{N}.$$
Averaging these two values gives the force at the midpoint: $93.3\, \text{N}$.

Question: in Part c), Why does the mass on the RHS of Newton's Second Law have to be the
rope plus the $5.00$-kg block?

 

[tiny-tim]

Hi Ackbeet!

c) What is the tension at the top of the heavy rope?
…
c) To find the tension in the rope at the top, we must sum the forces and
use Newton's Second Law. That is, we know that
$$F-mg-m_{5}g=(m+m_{5})a,$$
or
$$F=(m+m_{5})a+mg+m_{5}g=(9.00)(3.53)+(4.00)(9.8)+(5.00)(9.8)=120\, \text{N}.$$
…
Question: in Part c), Why does the mass on the RHS of Newton's Second Law have to be the
rope plus the $5.00$-kg block?

Because you are using F = ma on the rope-plus-lower-block.

You have to do this because you want the tension at the top of the rope, and so your body has to include the rope.

So F is all the external forces on the rope-plus-lower-block, and m is the mass of the rope-plus-lower-block.