Required force to start the motion of a leaning bar

This is a diagram of the reaction forces at points A and B. The reaction forces are perpendicular to the surface of contact for A. In the case of B, again the reaction force is perpendicular to a surface (think of it the reaction).

So you have the following forces:

• $N_A$: reaction force at point A
• $T_A$: Friction force at point A
• $N_B$: reaction force at point B $left(frac{N_{By}}{N_B{x}}=frac{3}{4}right)$
• $T_B$: Friction force at point B $left(frac{T_{By}}{T_B{x}}=frac{-4}{3}right)$

Regarding the friction forces they are drawn this way, because that will be direction of force when the force P is large enough for them to start moving. Additionally, the magnitude of the friction forces will be (just before movement is initiated):

$$T_A = mu N_A, quad T_B=mu N_B$$

In order to calculate $N_B$ we can estimate by the equilibrium of moments about point A:

$$sum M_A = 0 rightarrow N_Bcdot 5m - Wcdot left(4cdotfrac{3}{5}right) = 0$$

Where:

• W is the weight of the bar and its applier right at the middle of the beam (4[m] from the edge). The horizontal distance from point A is $left(4cdotfrac{3}{5}right)$

$$N_B = frac{12}{25}cdot W$$

The equilibrium of forces on Y axis is:

$$sum F_y = 0 rightarrow N_A + N_{By} -T_{By}-W = 0 $$ $$N_A + N_{B}frac{3}{5} - mucdot N_{B}cdot frac{4}{5} -W = 0 $$ $$N_A + frac{12}{25}cdot W(frac{3}{5}- mucdot frac{4}{5}) -W = 0 $$ $$N_A =left(1 + frac{12}{25}(mucdot frac{4}{5}-frac{3}{5} right)cdot W $$ $$N_A =frac{517}{625}cdot W $$

In order to calculate $P$ we can estimate by the equilibrium of moments about point B:

$$sum M_B = 0 rightarrow -N_Acdot 3m + (P- T_A)cdot 4m + Wcdot left(1cdotfrac{3}{5}right) = 0$$

$$ P=frac{1}{4}left(N_Acdot 3m - Wcdot left(1cdotfrac{3}{5}right) right) +T_A$$

$$ P=frac{1}{4}left(frac{517}{625}cdot Wcdot 3m - Wcdot left(1cdotfrac{3}{5}right) right) +mu cdot frac{517}{625}cdot W$$

$$ P=left(frac{1}{4}left(frac{517}{625}cdot 3 - frac{3}{5}right) +mu cdot frac{517}{625}right)cdot W$$

$$ P=frac{4491}{6250} cdot W$$