Question

Two 1.0 kg blocks are connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s2 by force F u .

a. What is F?
b. What is the tension at the top end of rope 1?
c. What is the tension at the bottom end of rope 1?
d. What is the tension at the top end of rope 2?

Answer 1

a)
`F=32.0 N`

b)
`T_(1)=19.2 N`

c)
`T_(2)= 16.0 N`

d)
`T_(3)=3.2 N`

Step-by-step explanation:

Let's do the free body diagram per each body. In this particular case the rope has a mass so we have:

First block

`F-T_(1)-m_(1)g=m_(1)a` (1)

Frist rope

`T_(1)-T_(2)-m_(r)g=m_(r)a` (2)

Second block

`T_(2)-T_(3)-m_(2)g=m_(2)a` (3)

Second rope

`T_(3)-m_(r)g=m_(r)a` (4)

When,

T

₁ is the tension at the top of the rope 1

T

₂ is the tension at the bottom of the rope 1

T

₃ is the tension at the top of the rope 2

Now, if we add all equations from (1) to (4), we will get the value of F,

`F-(m_(1)+m_(2)+2m_(r))g=(m_(1)+m_(2)+2m_(r))a`

a) Solving this equation for F:

`F=(m_(1)+m_(2)+2m_(r))(a+g)`

`F=(m_(1)+m_(2)+2m_(r))(a+g)`

`F=32.0 N`

b) Using the equation (1), we can find T₁

`T_(1)=F-m_(1)(g+a)`

`T_(1)=19.2 N`

c) Let's use the equation (2) to find T₂

`T_(2)=T_(1)-m_(r)(g+a)`

`T_(2)= 16.0 N`

d) Using the equation (4) we can find T₃:

`T_(3)=m_(r)(a+g)`

`T_(3)=3.2 N`

Have a nice day!