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

#### Chemist116

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
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

#### Chemist116

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

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