Question

A 8.57-m ladder with a mass of 21.4 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 258 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.63 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

`1311.5\ \text{Nm}`

Step-by-step explanation:

l = Length of ladder = 8.57 m

m = Mass of ladder = 21.4 kg

F = Force on ladder = 258 N

`\alpha` = Angular acceleration =
`1.63\ \text{rad/s}^2`

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

Net torque is given by

`\tau=lf-(l)/(2)mg\\\Rightarrow \tau=8.57* 258-(8.57)/(2)* 21.4* 9.81\\\Rightarrow \tau=1311.5\ \text{Nm}`

The net torque acting on the ladder is
`1311.5\ \text{Nm}`.