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

Jun 2017
399
6
Lima, Peru
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.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
$\tau_{net}$ about point A = 0 $\implies 3 \cdot F_B = 5 \cdot \dfrac{mg}{2} \implies F_B = \dfrac{5mg}{6}$

$\tau_{net}$ about point C = 0 $\implies 2 \cdot F_B = 5 \cdot F_A \implies F_A = \dfrac{2}{5} \cdot \dfrac{5mg}{6} = \dfrac{mg}{3}$

static_2.jpg
 
Jun 2017
399
6
Lima, Peru
$\tau_{net}$ about point A = 0 $\implies 3 \cdot F_B = 5 \cdot \dfrac{mg}{2} \implies F_B = \dfrac{5mg}{6}$

$\tau_{net}$ about point C = 0 $\implies 2 \cdot F_B = 5 \cdot F_A \implies F_A = \dfrac{2}{5} \cdot \dfrac{5mg}{6} = \dfrac{mg}{3}$

View attachment 10935
Why in this case the force on $A$ is pointing downwards, as it is the reaction, shouldn't it be upwards as it is on $B$?. Why the force acting on C and on the pivot in the ceiling are pointing upwards? and why is it $\frac{mg}{2}$? Why it looks as the tilt angle doesn't matter?. at both ends will it always be half of the mass of the object?.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Why in this case the force on A is pointing downwards, as it is the reaction, shouldn't it be upwards as it is on B?
First, consider the rotational equilibrium on the horizontal bar about point A. Clearly, the tilted bar is exerting a downward force at point C, hence there must be an upward force at point B for rotational equilibrium. Now, consider the rotational equilibrium about point C ... it has been established that the force at point B is upward, so what does that say about the direction of the force at point A?


Why the force acting on C and on the pivot in the ceiling are pointing upwards? and why is it mg/2? Why it looks as the tilt angle doesn't matter?. at both ends will it always be half of the mass of the object?
Reference the below diagram that shows the tilted bar with the pivot about its lower end ... what direction and magnitude would the force at the top have to be for the bar to be in a state of rotational equilibrium? Once you've answered that question, what direction and magnitude should the force at the lower end be for the bar to be in translational equilibrium in the vertical direction?

RotEq_tilt.jpg