# Equilibrium reaction forces on an inclined rod [closed]

Suppose we have a uniform steel rod leaning against a frictionless wall in static equilibrium. The frictional force between the lower end and the floor is less than its limiting value by a finite amount. The rod is supplied some amount of heat so that it expands. Assume that the coefficient of friction does not change on heating. Then if we assume the reaction force between rod and ground as $$R_1$$ and the reaction force between rod and wall as $$R_2$$, what would be the change in $$R_1$$ and $$R_2$$. According to our problem set, both should decrease.

## My Approach

I found from conservation in $$x$$ and $$y$$ directions that, $$R_2=f$$ where $$f$$ is the force of friction. And obviously $$R_1=mg$$. Now from applying the concept that net torque about any point would be zero, I ultimately arrive at the conclusion that $$mg=2R_{2}\tan\theta .$$ Now from intuition, I understand that if the rod expands, as there is no friction on the wall, it will tend to slide up thus the angle will increase, hence tan(Angle) should also increase, thus $$R_2$$ should also increase. But I cannot understand how to prove that $$R_1$$ will also increase. Is there something I am missing? All I know after this is that $$f\leq \mu R_{1}$$ But that isn't yielding anything useful.

Let's set up a coordinate system with $$+x$$ to the right, away from the wall, and $$+y$$ up from the ground. We can model the total force which the ground exerts on the rod as the sum of two components (we usually call these the "normal" force and friction). Because the wall is frictionless, we can see from a proper free-body diagram that the normal force component of the ground on the rod is $$\vec{F_n}=mg\hat{j}$$. Because the rod isn't sliding across the ground we also see that the friction of the ground on the rod is $$\vec{f}=R_2(-\hat{i})$$. Now we can calculate the total force the ground exerts on the rod: $$\vec{R}_1= -R_2\hat{i}+mg\hat{j} \\ R_1=|\vec{R}_1|=\sqrt{R_2^2+(mg)^2}$$

With the rod in a static position the sum of torques about any position must be zero. Again using a proper FBD, we chose the point of contact with the ground as the pivot point, calculate the torques (due to gravity at the center of mass and the wall force at the elevated end of the rod) and find that $$R_2=\frac{mg}{2\tan\theta}.$$

Now we see that, indeed, both $$R_1$$ and $$R_2$$ decrease when the angle increases.

Since there is no vertical force from the wall the vertical component of $$R_1$$, the normal reaction force $$N_1$$, must still match the weight and be unchanged. Then doing moments (sum of torques = 0) around the point of contact with the ground

$$R_2\times 2L \sin\theta = mg \times L \cos \theta$$

$$R_2 = \frac{mg}{2\tan\theta}$$

If the rod moves up the wall when expanding as you said, $$\theta$$ will increase, so $$R_2$$ decreases.

By resolving horizontally the friction force $$F_1$$ from the ground matches $$R_2$$ so $$F_1$$ decreases.

$$R_1 = \sqrt {N_1^2 + F_1^2}$$

and so $$R_1$$ also decreases.

• $R_1$ is not $mg$. The ground force has both horizontal (friction) and vertical (mg) components. Commented Nov 3, 2021 at 13:34
• @Bill N As the question called it a reaction force I presumed $R_1$ was the vertical force that matches the weight. You are right there is a frictional component $F_1$ say, since by resolving forces horizontally $F_1$ = $R_2$ (and $R_2$ decreased) we get the result that the total force at the ground decreases (by Pythagoras with the two components)...so it matches the book that both decrease. Commented Nov 3, 2021 at 20:04
• @ Bill N thankyou for clarifying, edit made. Commented Nov 3, 2021 at 20:14

As the rod expands due to heat, there will be relative motion with the ground and hence friction will increase. Therefore, the normal force on the wall will increase as $$f_r=R_2$$.

My approach:

on the wall $$\vec R_2 =\begin{pmatrix}R_{2x}\\0\end{pmatrix}$$ and on the floor $$\vec R_1 =\begin{pmatrix}R_{1x}\\R_{1y}\end{pmatrix}$$

sum of forces in x direction:

$$R_{2x}-R_{1x}=0$$ gives $$R_{2x} = R_{1x}$$

sum of forces in y direction:

$$-mg+R_{1y} = 0$$ gives $$R_{1y} = mg$$

further $$R_{2x}=\mu \, R_{1y} = \mu \, m g$$

moments on wall (counterclockwise):

$$\frac{-l\,mg\, cos\theta}{2} + l\, mg \, cos\theta - \mu\, l\, mg \, sin\theta =0$$

$$tan\theta = \frac{1}{2\mu}$$ for not sliding, we must postulate $$tan\theta > \frac{1}{2\mu}$$ otherwise our static assumption will not hold.

moments on floor (counterclockwise):

$$\frac{l\,mg\, cos\theta}{2} - l R_{2x}\, sin \theta = 0$$ it follows

$$R_{2x} = \frac{mg cos\theta}{2\,sin\theta}=\frac{mg}{2\, tan\theta} = \mu \,mg$$

finally we get

$$R_1 = \sqrt{R_{1x}^2+R_{1y}^2} = mg\, \sqrt{\frac{1}{4\, tan^2\theta} +1}\,$$ and

$$R_2 = \sqrt{R_{2x}^2} = mg\, \frac{1}{2\, tan\theta}$$

Now, it's obvious $$\theta>\theta'\, R_1 \frac{1}{2\mu}$$

• Your notation needs to be consistent with the question: $R_1$ refers to the floor (ground) force and $R_2$ refers to the wall. Commented Nov 4, 2021 at 2:09
• thanks I will correct Commented Nov 4, 2021 at 7:02