# How much force is needed to break off the stick

Let's consider the following figure

The grey box contains a blue stick which is fixed. The blue stick has a length of $$a+b+c$$ and two diameters $$f,h$$. The diameter $$h$$ describes the part $$b$$ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $$F$$ is pushing against the withe plane like in the picture. How much force is needed to break off the stick in part $$b$$?

• How much effort have you applied to try to obtain a proposed solution? Commented Feb 26, 2023 at 20:03
• I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane. Commented Feb 26, 2023 at 20:26
• I just have some knowledge about bending sticks and not about stuff like in my picture. Commented Feb 26, 2023 at 20:43
• Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. Commented Feb 26, 2023 at 20:57
• Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? Commented Feb 26, 2023 at 21:18

$$X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis. When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B.
$$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$
$$\sigma_y=\frac{MC}{I_{Beam}}=\frac{(F*x)(e+f/2)}{bh^3/12+bh(e+f/2)^2}$$ $$F*X=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{e+f/2}$$
$$F=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{(e*f/2)*(a+b+c+d/2)}$$