Question 4.7: 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 an unflanged 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.
Free-Body Diagram. A free-body diagram of the ladder is drawn. The forces involved are the combined weight of the man and ladder,
{ W}={-m{{g}}}~{\mathrm j}={-(\mathrm{80~kg+20~kg)}(9.81~{\ m/s^{2}})}{\mathrm j}={-(\mathrm{981~N})}{\mathrm j}and five unknown reaction components, two at each flanged wheel and one at the unflanged wheel. The ladder is thus only partially constrained; it is free to roll along the rails. It is, however, in equilibrium under the given load since the equation \Sigma F_{x}=0 is satisfied.
Equilibrium Equations. We express that the forces acting on the ladder form a system equivalent to zero:
Computing the vector products, we have \dagger
1.2B_{y}\mathrm{k}\,-\,1.2B_{z}\mathrm{j}\,-\,882.9\mathrm{k\,-\,588.6i\,-\,0.6Cj\,+\,3Ci\,=\,0} \\ \\ (3C\ -588.6){\mathrm i}\ -(1.2B_{z}+\ 0.6C){\mathrm j}+(1.2B_{y}-882.9){\mathrm k}=0 (2)
Setting the coefficients of i, j, k equal to zero in Eq. (2), we obtain the following three scalar equations, which express that the sum of the moments about each coordinate axis must be zero:
3C\ -\ 588.6=0\qquad\quad C=\ +196.2\ \mathrm{N} \\ \\ 1.2B_{z}\,+\,0.6C\,=\,0\qquad B_{z}\,=\,-98.1\,\,\mathrm{N} \\ \\ 1.2B_{y}-882.9=0~~~~~~~~B_{y}=~+736~\mathrm{N}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), we obtain two scalar equations expressing that the sums of the components in the y and z directions are zero. Substituting for B_{y},\,B_{z}, and C the values obtained above, we write
We conclude that the reaction at A is A = +(245 N)j – (98.1 N)k
\dagger The moments in this sample problem and in Sample Probs. 4.8 and 4.9 can also be expressed in the form of determinants (see Sample Prob. 3.10).
