# FBD with a system with glue

I am trying to understand how Adhesives or glues work. In the process I started thinking of drawing the Free body diagram. What will the FBD look like for the Blocks A, B and the adhesive itself? (I am unable to draw the same since to me the force by gravity and the glue both appears downward (for block A), which will not keep the block in equilibrium.)

The system is in equilibrium. B is resting on the ground.

• You need to explain your drawing. What if anything is supporting what I’m looking at? Are the two blocks and adhesive in free fall? Just what’s supposed to be going on here? Nov 28 '19 at 6:50

I believe you'll want to keep the entities (Block A, Adhesive, Block B) conceptually separate if you want to think about the free body diagrams. In particular, the free body diagram indicates the forces acting on the object. As you point out, Block A is in equilibrium (not accelerating), so the only force that can cancel gravity, which is pulling down, is a normal force upward that is coming from the Adhesive layer. Now, if you tried lifting Block A, then at this point the Adhesive would be the source of a downward force in addition to that of gravity. I hope this helps.

• You mean the nature of the adhesive will shift from pushing to pulling once we try to lift it? Nov 28 '19 at 9:43

Assuming block A is not moving and block B is hanging from it, (if it is accelerating, is it pulling block B or vice versa, but the concept would remain the same) we cut horizontally at the plane where the adhesive layer contacts the block B to spare ourselves worrying about the weight of the adhesive. And consider the FBD of the block a with the adhesive at this contact plane.

Let's call the sides of the adhesive a the long side facing front in the diagram and b the depth with the stipulation that block B is hanging balanced with block A and glue on side b, but as your diagram shows it is hanging off-center from the top by a distance D.

So the off-center moment causes tension stress at the right side of the contact surface and compression stress on the left side. these stresses distribute like an inclined line sloping down from the right hand to pass zero stress at the center and intercept compression stress which by our convention is negative. Let's call this maximum stress $$\ Sm= \sigma_{max-moment}$$

This triangular stress then adds or subtracts to the tension stress which is due to the weight of block B.

The two triangles have each the sides a/2 and Sm and area (read force) $$\ Ar= b*a/2*Sm/2= b*a*Sm/4$$ and the distance from one CG to the other will be 2/3*a as the couple arm.

$$\Sigma M= 0 \quad mB*g*D = 2/3a*(b*a*Sm/4)= 2/3a^2*Sm \\ Sm= \frac{mB*g*D}{b*2a^2/3}$$

And the total stress on the adhesive surface downward = $$\sigma_{adhesive}= \frac{mB*g}{a*b}\pm Sm.$$

If block A is accelerating we substitute $$\alpha\ for\ g$$ every place we see it, to get the motive force and moment.

If the side b of the adhesive contact surface is not balanced either we have to treat it the same way and add the two off-balance stresses as the square root of their sum of squares to the weight of block B stress.

• Ok , so is this a generalized expression, I mean even if the system is in resting on the ground ? Nov 28 '19 at 9:39
• If the system is at rest on the ground the weight of block A is supported by the adhesive and block B as compression stress but by your sketch balanced. Just divide mAg by ab. Because now the ground is handling the moment of A not sitting on the CG of B. Nov 28 '19 at 16:35