Question

A 8.89 -m ladder with a mass of 24.4 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 235 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.93 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? (b) What is the ladder's moment of inertia?

Answer 1

l=8.89m, m=24.4 kg, F_a=235N, α=1.93(rad/s²), θ=90^o, g=9.81(m/s²)

There are two forces acting on the ladder Gravity acting on the center of mass of the ladder, and the applied force. In order to obtain the net torque we must take the sum of the torques. Στ_net=Iα. Recall that τ=r×F=|r||F|sin(θ) so calculating the τ_G=|l/2||mg|sin(90)= |l/2||mg| Now the torque of the applied force |τ_a|=|l||F_a|sin(90)= |l||F_a|. |τ_G|=|8.89/2||(24.4)(9.81)|=1063.97 Nm and|τ_a|=|8.89||235|=2089.15 Nm

Now Στ_net=τ_a-τ_G (The torque produced by gravity is negative because its direction is opposite the direction of motion.)

Στ_net=2089.15-1063.97=1025.18 Nm using 3 significant figures τ_net=1030 Nm. Now recall that Στ_net=Iα ∴ (Στ_net/α)=I so (1025.18)/(1.93)=I so I=531.1813472 kgm2 or using 3 significant figures 531 kgm2

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