Question

Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.

Answer 1

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

$\mu_{sB}=0.126$

$\mu_{sC}=0.168$

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

$\sum \tau_{A}=0$

$N(3m)-W(1.5m)=0$

When solving for N we get:

$N=\frac{W(1.5m)}{3m}$

$N=\frac{(1962N)(1.5m)}{3m}$

$N=981N$

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

$\sum F_{y}=0$

$-F_{By}+N_{c}=0$

$F_{By}=N_{c}$

Next, the forces in x.

$\sum F_{x}=0$

$-f_{sB}-f_{sC}+P_{x}=0$

We can find the x-component of force P like this:

$P_{x}=360N(\frac{4}{5})=288N$

and finally the torques about C.

$\sum \tau_{C}=0$

$f_{sB}(1.75m)-P_{x}(0.75m)=0$

$f_{sB}=\frac{288N(0.75m)}{1.75m}$

$f_{sB}=123.43N$

With the static friction force in point B we can find the coefficient of static friction in B:

$\mu_{sB}=\frac{f_{sB}}{N}$

$\mu_{sB}=\frac{123.43N}{981N}$

$\mu_{sB}=0.126$

And now we can find the friction force in C.

$f_{sC}=P_{x}-f_{xB}$

$f_{sC}=288N-123.43N=164.57N$

$f_{sC}=N_{c}\mu_{sC}$

and now we can use this to find static friction coefficient in point C.

$\mu_{sC}=\frac{f_{sC}}{N}$

$\mu_{sC}=\frac{164.57N}{981N}$

$\mu_{sB}=0.168$