A 9.07-m ladder with a mass of 24.6 kg lies flat on the ground. A painter grabs

Question

A 9.07-m ladder with a mass of 24.6 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 1.99 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?

Explanation / Answer

l=8.89m, m=22.0kg, Fa=247N, =1.99(rad/s2), =90o, 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. net=I. Recall that =r×F=|r||F|sin() so calculating the G=|l/2||mg|sin(90)= |l/2||mg| Now the torque of the applied force |a|=|l||Fa|sin(90)= |l||Fa|. |G|=|8.89/2||(22.0)(9.81)|=959.3199Nm and|a|=|8.89||247|=2195.83Nm

Now net=a-G (The torque produced by gravity is negative because its direction is opposite the direction of motion.) net=2195.83-959.3199=1236.5101Nm using 3 significant figures net=1240Nm. Now recall that net=I (net/)=I so (1236.5101)/(1.99)=I so I=621.3618593kgm2 or using 3 significant figures 621kgm2

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