Romeo takes a uniform 10-m ladder and leans it against the smooth wall of the Capulet residence. The ladder's mass is 30kg, and the bottom rests on the ground 2.55m from the wall. When Romeo, whose mass is 70 kg, gets 88 percent of the way to the top, the ladder begins to slip. What is the coefficient of static friction between the ground and the ladder?(adsbygoogle = window.adsbygoogle || []).push({});

For this problem i attempted to find the torque of each of the masses (romeo and the ladder) add them together and with the equation f = u*N find the coefficient of static friction.

I found the torque by finding the angles of the ladder. I found that the angle the ladder makes with the ground to be 75.226. Therefore the torque of each of the masses will only be acting on the sin of 14.77 degrees. (this may be where i mess up but i don't know what else to do)

so i took these equations: mass(romeo)*9.81*sin(14.77)*distance from axis

mass(ladder)*9.81*sin(14.77)*distance from axis

70*9.81*sin(14.77)*8.8=torque of romeo = 1540

30*9.81*sin(14.77)*5= torque of ladder = 375

sum the torques to find the total force working against the ground to get 1915 N.

The normal force i found to be the mass of the total system times 9.81 so it is 981 N.

So i took 1915/981 to get the coefficient of static friction to be 1.95. this is not right, where did i mess up? thanks

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# Torque and static friction

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