required information. the 10 - m beam ab rests upon, but is not attached to, supports at c and d. given: f =…

Explanation:

Step1: Take moment about point C

For the beam to be in equilibrium, the sum of moments about point C must be zero. Let's consider the clock - wise and counter - clockwise moments. The moment of a force $F$ about a point is given by $M = F\times d$, where $d$ is the perpendicular distance from the point to the line of action of the force. $\sum M_C=0$. The forces are $P$ at a distance of $2$m from C, $F = 10$kN at a distance of $3$m from C and $20$kN at a distance of $(3 + 3+2)=8$m from C. $P\times2+F\times3-20\times8 = 0$ when considering the limiting cases for equilibrium.

Step2: Solve for the lower - bound of P

First, assume the beam is on the verge of moving in one direction. Let's solve the moment - equilibrium equation for $P$. $2P+10\times3 - 20\times8=0$ $2P+30 - 160 = 0$ $2P=160 - 30$ $2P = 130$ $P = 65$kN (this is one of the limiting values).

Step3: Consider the other limiting case

Now, assume the beam is on the verge of moving in the opposite direction. We can set up the moment - equilibrium equation with the appropriate signs for the moments. Let's take moments about point D. $\sum M_D=0$. The distance of $P$ from D is $(2 + 3)=5$m, the distance of $F$ from D is $3$m and the distance of $20$kN from D is $2$m. $P\times5-10\times3-20\times2 = 0$ $5P-30 - 40 = 0$ $5P=30 + 40$ $5P = 70$ $P = 14$kN

Answer:

$14$kN $\leq P\leq65$kN