Consider a ladder with a painter climbing up it. If the mass of the ladder is 12
ID: 1910797 • Letter: C
Question
Consider a ladder with a painter climbing up it. If the mass of the ladder is 12.0 kg, the mass of the painter is 68.0 kg, and the ladder begins to slip at its base when her feet are are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor? Assume the wall is frictionless.
Do not copy and paste a related answer from another question. I will rate badly. If you give me a correct answer, I will rate lifesaver.
Thanks!
Consider a ladder with a painter climbing up it. If the mass of the ladder is 12.0 kg, the mass of the painter is 68.0 kg, and the ladder begins to slip at its base when her feet are are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor? Assume the wall is frictionless. Do not copy and paste a related answer from another question. I will rate badly. If you give me a correct answer, I will rate lifesaver. Thanks!Explanation / Answer
How funny... I just gave my students a problem almost exactly like this one for their homework last week!
So you have to write three equations: horiz forces = 0 vert forces = 0 and torques = 0
horiz forces: W - f = 0 where W is force from wall and f is friction
vert forces: n - mg - Mg = 0 where n is normal force at ground
put these together... using f = un
W = un n = g(m+M) = 9.8*(12+68) = 784
Now for torques... we'll set the ref point at the lower end of the ladder. Then
torque from wall - torque from mg - torque from Mg = 0
W * 4 - mg*1.5 - Mg*2.1 = 0
4W = 1.5 * 12 * 9.8 + 2.1 * 68 * 9.8 = 1575.84
W = 393.96
finally
coeff of friction = u = W / n = 393.96 / 784 = 0.5025
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