Statics Practice Exam — Free

Replace the distributed load by its resultant, $R = 4.0 \times 3.0 = 12\,\text{kN}$, acting at the centroid of the loaded length, $x = 1.5\,\text{m}$. Draw the free-body diagram of the whole beam: pin reactions $A_x$ and $A_y$ at A, roller reaction $B_y$ at B, the 12 kN resultant at $x = 1.5$ m, and the 10 kN load at $x = 4.5$ m.

Take moments about A so that both unknown pin reactions drop out (counter-clockwise positive):

Vertical equilibrium then gives $A_y = 12 + 10 - 10.5 = 11.5\,\text{kN}$, and $A_x = 0$ because no horizontal loads act. Taking moments about B confirms the result: $A_y(6.0) = 12(4.5) + 10(1.5) = 69$, so $A_y = 11.5$ kN, and $A_y + B_y = 22$ kN equals the total applied load.

The tempting answer 13.5 kN comes from placing the 12 kN resultant at the end of the loaded region ($x = 3.0$ m) instead of at its centre; 11.5 kN is the reaction at A, not at B; and 22 kN is the total load, which the two supports share.

$B_y = 10.5\,\text{kN}$ upward

Support reactions first. The truss and the load are symmetric about $x = 3$ m, so each support carries half the load: $A_y = B_y = 6\,\text{kN}$ upward, and $A_x = 0$ because there is no horizontal load.

Member AC runs from $(0,0)$ to $(3,4)$, so its length is $\sqrt{3^2 + 4^2} = 5$ m and its direction cosines are $3/5$ (horizontal) and $4/5$ (vertical).

Isolate joint A. Assume both member forces $F_{AB}$ and $F_{AC}$ are tensile (pulling away from the joint). Vertical equilibrium involves only $F_{AC}$ and the reaction, so it is solved first:

The negative sign means AC is in compression (7.5 kN), as expected for a top chord pushing down on the support. Horizontal equilibrium then gives the bottom chord:

The positive result confirms the tension assumption: the inclined members push the supports apart and the bottom chord ties them together. By symmetry $F_{BC} = -7.5$ kN as well, and joint C closes: the two vertical components $2(7.5)(4/5) = 12$ kN balance the applied load.

$F_{AB} = 4.5\,\text{kN}$, tension

Free-body diagram of the ladder: its weight $W_L = 20(9.81) = 196.2$ N at the midpoint (2.5 m along the ladder), the person's weight $W_P = 70(9.81) = 686.7$ N at a distance $d$ along the ladder, a horizontal normal force $N_w$ from the smooth wall at the top, and at the base a vertical normal force $N_f$ plus a horizontal friction force $f$ directed toward the wall (friction opposes the base sliding away from the wall).

Vertical equilibrium gives $N_f = 196.2 + 686.7 = 882.9$ N. Horizontal equilibrium gives $f = N_w$.

At impending slip the floor friction is at its maximum, $f = \mu_s N_f = 0.35(882.9) = 309.0$ N, so $N_w = 309.0$ N as well.

Take moments about the base so that $N_f$ and $f$ drop out. The wall force has lever arm $L\sin 60^\circ$, while each weight has a horizontal lever arm equal to its distance along the ladder times $\cos 60^\circ$:

With nobody on the ladder the friction required is only $196.2(2.5)(0.5)/(5.0 \times 0.8660) = 56.6$ N, far below the 309 N available, so the ladder is stable on its own and only the person's added moment uses up the remaining friction.

The person can climb 3.18 m along the ladder (about 64% of its length) before it slips.