Question 4.SP.7: A 20-kg ladder used to reach high shelves in a storeroom is ...
A 20-kg ladder used to reach high shelves in a storeroom is supported by two flanged wheels A and B mounted on a rail and by a flangeless wheel C resting against a rail fixed to the wall. An 80-kg man stands on the ladder and leans to the right. The line of action of the combined weight W of the man and ladder intersects the floor at point D. Determine the reactions at A, B, and C.
STRATEGY: Draw a free-body diagram of the ladder, and then write and solve the equilibrium equations in three dimensions.
MODELING:
Free-Body Diagram. The combined weight of the man and ladder is
W = −mgj = −(80 kg + 20 kg)(9.81 m/s²)j = −(981 N)j
You have five unknown reaction components: two at each flanged wheel and one at the flangeless wheel (Fig. 1). The ladder is thus only partially constrained; it is free to roll along the rails. It is, however, in equilibrium under the given load because the equation \Sigma F_x = 0 is satisfied.
ANALYSIS:
Equilibrium Equations. The forces acting on the ladder form a system equivalent to zero:
\begin{gathered}\Sigma \pmb{F}=0: \quad A_y \pmb{j}+A_z \pmb{k}+B_y \pmb{j}+B_z \pmb{k}-(981 N) \pmb{j}+C \pmb{k}=0 \\\left(A_y+B_y-981 N\right) \pmb{j}+\left(A_z+B_z+C\right) \pmb{k}=0\end{gathered} (1)
\Sigma \pmb{M}_A=\Sigma(\pmb{r} \times \pmb{F})=0: \quad 1.2 \pmb{i} \times\left(B_y \pmb{j}+B_z \pmb{k}\right)+(0.9 \pmb{i}-0.6 \pmb{k}) \times(-981 \pmb{j})+(0.6 \pmb{i}+3 \pmb{j}-1.2 \pmb{k}) \times CComputing the vector products gives you { }^{\dagger}
\begin{array}{r}1.2 B_y \pmb{k}-1.2 B_z \pmb{j}-882.9 \pmb{k}-588.6 \pmb{i}-0.6 C \pmb{j}+3 C \pmb{i}=0 \\ (3 C-588.6) \pmb{i}-\left(1.2 B_z+0.6 C\right) \pmb{j}+\left(1.2 B_y-882.9\right) \pmb{k}=0\end{array} (2)
Setting the coefficients of i, j, and k equal to zero in Eq. (2) produces the following three scalar equations, which state that the sum of the moments about each coordinate axis must be zero:
\begin{array}{rlrl}3 C-588.6 & =0 & & C=+196.2 N \\1.2 B_z+0.6 C & =0 & & B_z=-98.1 N \\1.2 B_y-882.9 & =0 & & B_y =+736 N\end{array}The reactions at B and C are therefore
B = +(736 N)j −(98.1 N)k C = +(196.2 N)k
Setting the coefficients of j and k equal to zero in Eq. (1), you obtain two scalar equations stating that the sums of the components in the y and z directions are zero. Substitute the values above for B_y, B_z, and C to get
\begin{array}{rrrl}A_y+B_y-981=0 & A_y+736-981 & =0 & A_y=+245 N \\A_z+B_z+C=0 & A_z-98.1+196.2 & =0 & A_z=-98.1 N\end{array}Therefore, the reaction at A is
A = +(245 N)j −(98.1 N)k
REFLECT and THINK: You summed moments about A as part of the analysis. As a check, you could now use these results and demonstrate that the sum of moments about any other point, such as point B, is also zero.
{ }^{\dagger}The moments in this sample problem, as well as in Sample Probs. 4.8 and 4.9, also can be expressed as determinants (see Sample Prob. 3.10).
