Question

While using a pencil sharpener, a student applies the forces and couple shown.

(a) Determine the forces exerted at \(\mathrm{B}\) and \(\mathrm{C}\) knowing that these forces and the couple are equivalent to \(\mathrm{a}\)

force couple system at A consisting of the force \(\mathbf{R}=(2.6 \mathrm{lb}) \mathrm{i}+R_{yj} -(0.7 \mathrm{lb}) \mathbf{k}\) and the couple \(M_{A}^{R}=M_{x} \mathbf{i}+\)\(\left(1.0 lb* f t\right) j-\left(0.72 l b* f t\right) \mathbf{k} .\)

(b) Find the corresponding values of \(R_{y}\) and \(M_{x}\).

Answer 1

Given

$$ \begin{array}{l} \mathbf{R}=(2.6 \mathrm{lb}) \mathbf{i}+R_{y} \mathbf{j}-(0.7 \mathrm{lb}) \mathbf{k} \\ \mathbf{M}_{A}^{R}=M_{x} \mathbf{i}+(1.0 \mathrm{lb} \cdot \mathrm{ft}) \mathbf{j}-(0.72 \mathrm{lb} \cdot \mathrm{ft}) \mathbf{k} \end{array} $$

Figure:

We know that Resultant force, \(\mathbf{R}=\mathbf{B}+\mathbf{C}\) Sum of the forces al ong \(x\) axis \(\sum F_{x}=0\)

\(B_{x}+C_{x}=2.6 \mathrm{lb} \dots (1)\)

Similarly, Sum of the forces along \(y\) axis \(\sum F_{y}=0\)

\(-C_{y}=R_{y} \dots (2)\)

Now, Sum of the forces along \(z\) axis \(\sum F_{z}=0\)

\(-C_{z}=-0.7 \mathrm{lb}\)

\(C_{z}=0.71 \mathrm{~b}\)

Taking moments about \(x\) axis

\(\sum M_{x}=0\)

\(M_{x}=1+\left(\frac{1.75}{12}\right) C_{y} \dots (3)\)

Taking moments about \(y\) axis \(\sum M_{y}=0\)

\(\left(\frac{3.75}{12}\right) B_{x}+\left(\frac{1.75}{12}\right) C_{x}+\left(\frac{3.5}{12}\right)(0.71 \mathrm{~b})=1\)

\((3.75 \mathrm{ft}) B_{x}+(1.75 \mathrm{ft}) C_{x}=9.55 \mathrm{lb} \cdot \mathrm{ft}\)

Now from equation (1), we get \(C_{x}=2.6 \mathrm{lb}-B_{x}\)

Now, substituting the expression for \(C_{x}\) \((3.75 \mathrm{ft}) B_{x}+(1.75 \mathrm{ft})\left(2.6 \mathrm{lb}-B_{x}\right)=9.55 \mathrm{lb} \cdot \mathrm{ft}\)

\((2 \mathrm{ft}) B_{x}=5 \mathrm{lb} \cdot \mathrm{ft}\)

\(B_{x}=\frac{5 \mathrm{lb} \cdot \mathrm{ft}}{2 \mathrm{ft}}\)

\(B_{x}=2.5 \mathrm{lb}\)

\(B=(2.51 b)\) i

Now, we know that \(C_{x}=2.6 \mathrm{lb}-B_{x}\)

\(C_{x}=2.6 \mathrm{lb}-2.5 \mathrm{lb}\)

\(C_{x}=0.1 \mathrm{lb}\)

Taking moments about \(z\) axis \(\sum M_{z}=0\)

\(-\left(\frac{3.5}{12} \mathrm{ft}\right) C_{y}=-0.72 \mathrm{lb} \cdot \mathrm{ft}\)

\(C_{y}=2.47 \mathrm{lb}\)

\(\mathrm{C}=(0.11 \mathrm{~b}) \mathrm{i}-(2.47 \mathrm{lb}) \mathrm{j}-(0.71 \mathrm{~b}) \mathrm{k}\)