How to find the reaction about a fixed point between two bars when the tilt angle of one isn't known?

Question

The problem is as follows:

A horizontal bar is frictionless and its mass is negligible. Find the reaction at point $A$ and on point $B$ if the other bar is homogeneous and of $2.4\,kg$. Assume $g=10\frac{m}{s^2}$.

The alternatives given are:

$\begin{array}{ll} 1.&\textrm{8N and 20N}\\ 2.&\textrm{10N and 8N}\\ 3.&\textrm{8N and 15N}\\ 4.&\textrm{6N and 15N}\\ 5.&\textrm{6N and 20N}\\ \end{array}$

I'm confused exactly on this problem. How am I supposed to find the reaction at the requested point if not information is given concerning the tilt angle of the second bar?. The one which is held to a support from the ceiling?. Can someone help me with this please?. My book says the answer is the first option. But I have no idea how to get there.

Answer 1

Hint: since the contact between the two bars is without friction, there is no horizontal component at their contact point. There is only a vertical component which for the equilibrium of the vertical bar will be..

half of the weight of the swinging bar ..

--- reply to your comment ---

I think I caught your perplexity and I will try to clear it.

Take for simplicity two 2D rigid bodies. They can mutually interact only through the points of contact.

At those points (actually of course in a small area around it) they can exchange two type of forces:
- one reacting againts compenetration;
- one arising from small irregularities, and/or superficial adherence, collectively denominated friction, which will react against the two bodies sliding, or even detaching from (and thus rotating over) each other.

Assume, for simplicity again, that the point of contact be just one. If the friction is negligible then the reaction against compenetration will be essentially normal to the small area of contact, i.e. it will be normal to the common tangent line there, which also means that is directed along the line connecting the centers of curvature at the contact point.

Now, the horizontal bar is flat and the inclined bar has a rounded tip which ensures that the contact area be limited and so with negligible friction. Then it is clear that the reaction at that point will be just vertical.

Answer 2

Gravity exerts a downward force on the tilted bar which produces a torque around the pivot point. You can measure that torque by taking the weight of the bar and supposing the weight is a downward force exerted at the center of the bar. The torque is the force times the horizontal distance from the center of the bar to the pivot.

The horizontal bar can only exert a force directly upward on the tilted bar. That force produces a torque equal to the force times the horizontal distance from the end of the bar to the pivot.

Set up an equation in which the two torques cancel out and figure out the unknown force.

---

Possibly a simpler way to look at it:

A bar of mass $2.4$ kg is supported by an upward force at each end of the bar. Find the two forces if the bar is not moving and is not vertical.

If the bar were vertical you could apply more or less force at the top end (and correspondingly less or more force at the bottom) without disturbing the equilibrium. But when the bar is tilted there is only one answer.

---

Another way to look at it:

Since the angle of the tilted bar is a source of confusion, let's look just at that bar and what we could do with it. The figure below shows a few examples.

In case A, the bar is hanging straight down from the pivot at the upper end and no other object is acting on it. So the only forces on the bar are the force of gravity (actually acting on all parts of the bar, but simplified here to a single force applied to the center of mass) and the force exerted on the top end of the bar where it is attached to a pivot. Gravity is straight down, so for the components of the two forces to cancel out (so that the bar does not move), the force at the pivot must be straight up and equal to the weight of the bar.

In case B, the bar is tilted. There are still no other objects acting on it except for the pivot. Is it possible for the bar to be in equilibrium?

In order for case B to be in equilibrium, the forces have to cancel out in both the $x$ and $y$ directions. This means the force at the pivot must still be straight up. But now we have two forces acting in opposite directions along different lines of action. This would cause the bar to turn counterclockwise. So no, it is not possible for the bar to be in equilibrium in case B.

The only equilibrium position for a bar attached to a pivot like this, with no other forces acting on it, is vertical.

In case C, someone's finger is pushing against the side of the bar to hold it at a non-vertical angle. Assuming the finger is pushing perpendicular to the side of the bar (which is the only direction it could push if the bar's surface is frictionless), the angle of the force depends on the angle of the bar, and therefore so does its decomposition into horizontal and vertical forces. If the finger is holding the bar motionless in this position, equilibrium requires the pivot to exert a horizontal force to balance the horizontal component of the finger's force.

The thing is, if you think about a bar being supported in equilibrium at a tilted angle, case C is what your mind would tend to imagine, because that is how we would most naturally put a bar into such a tilted equilibrium state: push against the side of the bar until it reaches the desired angle, and hold it there. And indeed the angle of the bar matters very much in this case. It is also a relatively complicated problem to balance the forces so that the bar is not accelerated in any direction or rotated.

In case D we have placed a horizontal frictionless surface under the lower end of the bar. The force exerted by that surface has to be perpendicular to the surface (since it is frictionless), and therefore has to be straight up regardless of the angle of the bar. Since the force of gravity and the force at the bottom have no horizontal component, if we want the bar to be in equilibrium the force at the pivot cannot have any horizontal force either.

So now we have three forces acting at three different points on the bar, all of them purely vertical. In order to have equilibrium, the forces must be balanced so they do not rotate the bar (unlike case B).

It comes down to measuring the distances between the lines of action of the three forces, and making sure the moments cancel out. The magnitudes of the two forces at the ends have to add up to the magnitude of the force of gravity, so there's only one thing you can vary, namely, what percentage of the upward force is at the bottom rather than the top. Because the setup is so symmetrical (the downward force on a line exactly halfway between the other two lines of force), the two upward forces turn out to be equal. But you can work this out formally by setting up an equation of the moments of the forces around some point.

The thing is to remember that when you have the horizontal bar holding up the lower end of the tilted bar, it is case D and not case C.

Answer 3

As the horizontal bar is frictionless, the action and reaction between the two bars is vertical. The oblique bar acts as a lever, hence the force intensity is half the weight (which is applied to the middle point).

At $A$, $\dfrac23\cdot10\cdot\dfrac{2.4}2=8$, at $B$, $8+10\cdot\dfrac{2.4}2=20$.

Answer 4

Based on the figure and the text of the problem, the static scheme turns out to be the following:

Hence, it's sufficient to remove the constraints and introduce the respective constraint reactions:

All that remains is to impose the static equilibrium of both bars:

$$ \begin{cases} H_A = 0 \\ V_A + V_B + V_C = 0 \\ V_B\,L_1 + V_C\left(L_1+L_2\right) = 0 \\ . \\ H_E = 0 \\ - V_C - P + V_E = 0 \\ V_C\,(2\,L_3) + P\,L_3 = 0 \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} H_A = 0 \\ V_A = - \frac{L_2}{L_1}\,\frac{P}{2} \\ V_B = \frac{L_1 + L_2}{L_1}\,\frac{P}{2} \\ V_C = - \frac{P}{2} \\ H_E = 0 \\ V_E = \frac{P}{2} \end{cases} $$

where the minus signs indicate that the respective verses previously hypothesized are contrary to the real ones. In conclusion, the resulting constraint reactions required are equal to:

$$ R_A = \sqrt{H_A^2 + V_A^2} = \frac{L_2}{L_1}\,\frac{P}{2} = \frac{2\,m}{3\,m}\,\frac{24\,N}{2} = 8\,N \,; \\ R_B = \left|V_B\right| = \frac{L_1 + L_2}{L_1}\,\frac{P}{2} = \frac{5\,m}{3\,m}\,\frac{24\,N}{2} = 20\,N \,; $$

hence the correct answer is A. All this to show the standard procedure applicable in each case (at least in the isostatic structures, where the hypothesis of rigid body is sufficient), obviously once learned in simple cases like this you can solve everything by eye.