Question

A chain consists of 6 links, each having a mass of 0.117 kg. The chain is lifted vertically by applying a force to the top link with an upward acceleration of 2.13 m/s 2 . No point on the chain touches the floor. acosta (baa2565) – Dynamics – shaikh – (PHYS-2425) 2 The acceleration of gravity is 9.81 m/s 2 . Find the force exerted on the top of the chain. Answer in units of N.

Answer 1

The force exerted on the top of the chain is 7.4316 N.

Step-by-step explanation:

Given that,

Mass of link = 0.117 kg

Number of links = 6

Acceleration = 2.13 m/s²

We need to calculate the force on first link due to second link

Using formula of force

`F=m(g+a)`

Put the value into the formula

`F_(1)=0.117(9.81+2.13)`

`F_(1)=1.3969\ N`

The force on second link due to third link

`F_(2)=m(g+a)F_(1)`

Put the value into the formula

`F_(2)=0.117(9.81+2.13)*1.3969`

`F_(2)=1.9514\ N`

The force on third link due to four link

`F_(3)=m(g+a)F_(2)`

Put the value into the formula

`F_(3)=0.117(9.81+2.13)*1.9514`

`F_(3)=2.7260\ N`

The force on fourth link due to five link

`F_(4)=m(g+a)F_(3)`

Put the value into the formula

`F_(4)=0.117(9.81+2.13)*2.7260`

`F_(4)=3.8081\ N`

The force on fifth link due to six link

`F_(5)=m(g+a)F_(4)`

Put the value into the formula

`F_(5)=0.117(9.81+2.13)*3.8081`

`F_(5)=5.3198\ N`

We need to calculate the force exerted on the top of the chain

`F_(6)=m(g+a)F_(5)`

Put the value into the formula

`F_(6)=0.117(9.81+2.13)*5.3198`

`F_(6)=7.4316\ N`

Hence, The force exerted on the top of the chain is 7.4316 N.