How to find the modulus of the force that the floor excerts to a circular plate connected to a bar?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

The figure from below shows a bar which mass is $3\,kg$ and a circular plate which its mass is $5\,kg$. Assume that both are homogeneous. Find the modulus of the force that the floor excerts to the circular plate. Assume that the acceleration due gravity is $10$ meters per second square and point $O$ is a screw which joins the bar with the circular plate.



The alternatives are as follows:

$\begin{array}{ll}
1.&15N\\
2.&30N\\
3.&65N\\
4.&50N\\
5.&60N\\
\end{array}$

In this problem, I'm lost as to exactly how I should take the interpretation of the forces acting on the body. Can someone help me with the FBD for this object please? At first, I assumed that since there are $3$ and $5$ kilograms then the reaction would be 15N upwards. But I'm not sure about this answer. Help please?
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Left side shows forces acting on the bar. Right side shows forces acting on the plate.

I get the Normal force the floor exerts on the plate to be 65N

bar_disk.jpg
 
Jun 2017
399
6
Lima, Peru
Left side shows forces acting on the bar. Right side shows forces acting on the plate.

I get the Normal force the floor exerts on the plate to be 65N

View attachment 10936
Is the angle $\theta$ relevant for solving this problem?. What system of equations did you used to get that value for the reaction?.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
torques about the lower right end of the bar ...

$\displaystyle \sum \tau = 0 \implies mg\sin{\theta} \cdot \dfrac{L}{2} = R_{db}\sin(180-\theta) \cdot L \implies \dfrac{mg}{2} = R_{db}$

note ${\color{red}R_{db}} = {\color{blue}R_{bd}}$

$N = R_{bd} + Mg = \dfrac{mg}{2} + Mg$