Question

If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor

Answer 1

Answer:

μ = 0.498

Explanation:

I manage to find the picture of this problem. In the attached picture you can also see the forces involved in this case.

We know according to newton's law that the static friction force is:

Fs = μ * N

However, as you can see in the picture, the Normal force is equals to the weight, in this case the weight of the painter and also the ladder, therefore:

N = Fy

and the force of the friction is:

Fs = Fx

Therefore the coefficient is:

μ = Fx/Fy  (1)

Now, let's write the equations in x and y, to solve this.

For the "x" axis:

Fx - Fw = 0 -----> Fx = Fw   (2)

Fw is force of the wall, while Fx is the force friction in the x axis (base of the ladder).

for the "y" axis:

Fy - W1 - W2 = 0

W1 = mg (ladder)

W2 = Mg (painter)

replacing we have:

Fy = W1 + W2

Fy = mg + Mg ----> Fy = g(m + M)    (3)

To get the force that the wall is exerting we need to calculate the torque around the foot of the ladder so:

τ = rF sinθ

However, the angle in the wall and the ladder is 90° so:

τ = rF sin90°

τ = rF   (4)

replacing (4) with the forces we have:

rFw = rW1 + rW2

4Fw = 1.5mg + 2.1Mg

Fw = 1.5mg + 2.1Mg/4   (5)

Finally, with this expression, we can replace it in (1) to get the coefficient of friction:

μ = Fx/Fy

μ = 1.5mg + 2.1Mg / 4g(m + M)     gravity cancels out so:

μ = 1.5m + 2.1M / 4(m + M)

Replacing the data we finally have:

μ = (1.5 * 2) + (2.1 * 55) / 4 (55 + 12)

μ = 133.5 / 268

μ = 0.498

Answer 2

The coefficient of static friction between the ladder and the floor is 0.5.

From the image:

$n=F_{GY},F_s^{max}=F_{GX}\\\\F_s^{max}=\mu_sn\\\\F_{GX}=\mu_sF_{GY}\\\\sum \ of\ horizontal\ force\ is\ 0:F_{GX}-F_W=0;F_{GX}=F_W\\\\sum \ of\ vertical\ force\ is\ 0:F_{GY}-mg-Mg=0;F_{GY}=g(m+M)\\\\\\Total\ torque\ at\ point\ A(foot\ of\ ladder)=0, hence:\\\\mg(1.5)+Mg(2.1)-F_W(4)=0\\\\mg(1.5)+Mg(2.1)=F_W(4)\\\\mg(1.5)+Mg(2.1)=F_{GX}(4)\\\\mg(1.5)+Mg(2.1)=\mu_sF_{GY}(4)\\\\mg(1.5)+Mg(2.1)=\mu_sg(m+M)(4)\\\\\mu_s=\frac{mg(1.5)+Mg(2.1)}{4g(m+M)}$

$\mu_s=\frac{12(1.5)+55(2.1)}{4(12+55)} \\\\\mu_s=0.5$

Hence, the coefficient of static friction between the ladder and the floor is 0.5.

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