# Where to put weight so that 2 jointed rods are most likely to slip? [closed]

I'm doing the attached mechanics question for school and would like to know exactly why you the man standing in the middle of the system increases the likelihood of the system slipping. The solution says something quite hand wavy about the system being symetrical. I have tried to play around with the equations but have struggled with the unknown distance. Any help is much appreciated. I did my calculations but I don't get the same answer. Look at my approach, maybe it'll get you some idea. The answer does not seem to be correct.

Let's look at the AB rod only and lets name the angle $$\theta$$ such that $$cos(\theta) = d/l$$.

In horizontal (x) direction you can write an equation:

$$\mu F_N = F \tag{1}$$ where $$F$$ is the force of AC rod acting on AB rod.

In vertical (y) direction:

$$mg + Mg = F_N \tag{2}$$ where $$M$$ is a mass of a person.

And you need third equation for momentum (let's say about point B):

$$Fl\sin(\theta) - mg\frac{l}{2}\cos(\theta) - Mgx\cos(\theta) = 0 \tag{3}$$

With little trigonometry you can get that $$\sin(\theta) = \sqrt{l^2 - d^2}/{l}$$

Substituting (1) and (2) into (3) you get:

$$\mu (m+M)gl\sin(\theta) - mg\frac{l}{2}\cos(\theta) - Mgx\cos(\theta) = 0$$

$$\mu = \frac{mg\frac{l}{2} + Mgx}{(m+M)gl\tan(\theta)}$$

$$\mu = \frac{(mg\frac{l}{2} + Mgx)d}{(m+M)gl\ \sqrt{l^2 - d^2}}\tag{4}$$

You are looking for minimal value of $$\mu$$ for which (4) holds for every x $$\in [0, l]$$. That value you get for x=l because there the $$torque$$ of the person is maximized.

You get:

$$\mu_\text{min} = \mu(x=l) = \frac{(m+2M)d}{2(m+M)\sqrt{l^2 - d^2}}$$

Since ladder and a person have the same weight, you get:

$$\mu_\text{min} = \frac{3d}{4\sqrt{l^2 - d^2}}$$

• Nicely done, but in a static situation we don't need momentum. – R.W. Bird Oct 7 '19 at 19:15
• I am sorry for bad naming... Where I said moment, I meant torque because in my language torque is named "moment", therefore the confusion. I edited the post, thanks. – Luka Mandić Oct 7 '19 at 19:39
• Sorry for the late response, for the second equation do you not have to take into account a vertical force from the rod AC on the rod AB. – Greevman Oct 12 '19 at 11:07
• there is no vertical force from AC to rod AB. You can see it like this. Let's say that there is a force from AC to AB in downward direction. By Newton's 3rd law, there should be a force from AB to AC in upward direction. By the symmetry of the problem, that cannot be true and therefore there is no vertical force. – Luka Mandić Oct 13 '19 at 11:40
• Sorry, if the man is on the right side of the ladder, there is a vertical component of force from the left side. – R.W. Bird Oct 15 '19 at 14:29

We are looking for a coefficient that can handle the maximum required friction. Considering only the right side ladder, putting the man at the top will maximize the CCW torque around point C. To balance this we need the maximum horizontal force from the left side ladder. The friction must equal this force. (Under conditions of symmetry, neither ladder can be pushing the other up or down.) (With “l” as the length of the ladder, the square root is the height, h..) Taking torques about the top or bottom of the ladder, I get a required coefficient of (¾) d / h.

• I have made a mistake. I forgot that for symmetry the man at the top must put half of his weight on each of the ladders. That changes the torque arund the lower end of the ladder and the normal force. The friction becomes Wd/h and the minimum coefficient becomes (Wd/h)/(3W/2) = (2d)/(3h). If the man puts all his weight on the right side ladder, then the force from the left side has an upward vertical component. Taking this into consideration requires torque equations from both ladders and yields the same result. – R.W. Bird Oct 8 '19 at 16:40