# Equation to model change in leverage point

Is there a mathematical model for the situation below?

First, I want to know how much force I would need to lift this block some distance using this wooden board. The block is situated against a wall and also cannot horizontally fall. By lifting the board up, this also lifts the block up. The board is frictionless and the base of the board is fixed to the ground ($\mu_{fw} = 0$; $\mu_{fb} = 1$).

Then what is the force needed as the base of the board changes? I would like to graph this, so taking two discrete points will not suffice.

For example, as I'm lifting the board its base naturally starts slipping away from the wall ($\mu_{fw} > 0$; $\mu_{fb} < 1$).

• Is there friction between the block and the wall? Have you tried drawing free body diagrams (FBDs) for the block and the lever? Because the contact between the lever and block is frictionless the forces between them are normal (perpendicular) to the lever. Only the vertical component of this normal force $N$ raises the block, working against the weight of the block and any friction at the wall. The horizontal component of $N$ affects the amount of friction at the wall. As the angle of the board increases, the force required at the end of the lever also increases. – sammy gerbil Apr 17 '18 at 11:43
• @sammygerbil I would accept this as an answer. – Jossie Calderon Apr 21 '18 at 0:52

The starting point is to draw Free Body Diagrams (FBDs) for the board and the block.

Block:

Assuming there is no friction between block and walls, the horizontal forces on the block do not affect the vertical forces. So we can ignore the horizontal forces.

The only vertical forces on the block are its weight $W$ and the vertical component $N_1 \sin\theta$ of the normal reaction $N_1$ between the board and block, where $\theta$ is the angle which the board makes with the vertical. If the block is lifted with constant speed so that it does not accelerate, these 2 forces are equal at all times.

Board:

The board AB is a lever. I will assume that its base remains fixed to the ground at A, and it makes contact with the block at P. We do not need to consider reaction forces at A, because this is where the board pivots, and the reaction forces exert no moment about this point.

Just as the board does not accelerate upwards, so also it does not accelerate as it rotates. So the clockwise and anticlockwise moments of forces acting on it about any point are equal : $$N_1 p=FL\sin\theta$$ where $L=AB$ is the length of the board and $p=AP$.

The distance of P from the wall is always a fixed distance $X=a+p\sin\theta$ equal to the horizontal width of the block, where $a \le X$ is the fixed distance of A from the wall. The vertical distance of the block above the ground is $y=p\cos\theta$.

This should give you enough equations to calculate $F$ as a function of $\theta$ or as a function of $y$.

If the board is not fixed at A and is able to slip away from the wall as the angle $\theta$ increases, this is a more difficult problem.

Now $a$ depends on $\theta$ and the coefficient of friction $\mu$ between board and ground. The normal reaction at the ground is $N_2=W-F$ and the maximum static friction force is $R=\mu N_2$. The rod will not slip away from the wall until $N_1 \cos\theta = R = \mu N_2$. Thereafter it will slip as $\theta$ increases, and this condition will continue to be met as it does so.

At each value of $\theta$ the board is in quasi-static equilibrium, so we can continue to use the fact that the resultant moment of forces about A is zero.