Question

A 9.83 -m ladder with a mass of 20.3 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 242 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.64 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder

Answer 1

The net torque will be "1366.33 Nm".

Step-by-step explanation:

The given values are:

Length of ladder,

`\tau_(242N) = 9.83 \ m`

then,

`\tau=4.92`

Mass,

= 20.3 kg

Now,

The net torque will be:

⇒
`\tau_(net)=\tau_(mg)+\tau_(242)`

On putting the given values, we get

⇒
`=4.92* (-mg)+9.83* (+242)`

⇒
`=4.92* (-21* 9.8)+9.83* 242`

⇒
`=-1012.53+2378.86`

⇒
`=1366.33 \ Nm`