Question 3.SP.5: A cube of side a is acted upon by a force P along the diagon......
A cube of side a is acted upon by a force P along the diagonal of a face, as shown. Determine the moment of P (a) about A, (b) about the edge AB, (c) about the diagonal AG of the cube. (d) Using the result of part c, determine the perpendicular distance between AG and FC.
STRATEGY: Use the equations presented in this section to compute the moments asked for. You can find the distance between AG and FC from the expression for the moment M_{A G}.
Final Answer
MODELING and ANALYSIS:
a. Moment about A. Choosing x, y, and z axes as shown (Fig. 1), resolve into rectangular components the force P and the vector \mathbf{r}_{F / A}=\overrightarrow{A F} drawn from A to the point of application F of P.
\mathbf{r}_{F / A}=a \mathbf{i}-a \mathbf{j}=a(\mathbf{i}-\mathbf{j})
\mathbf{P}=(P / \sqrt{2}) \mathbf{j}-(P / \sqrt{2}) \mathbf{k}=(P / \sqrt{2})(\mathbf{j}-\mathbf{k})
The moment of P about A is the vector product of these two vectors:
\mathbf{M}_A=\mathbf{r}_{F / A} \times \mathbf{P}=a(\mathbf{i}-\mathbf{j}) \times(P / \sqrt{2})(\mathbf{j}-\mathbf{k})
\mathbf{M}_A=(a P / \sqrt{2})(\mathbf{i}+\mathbf{j}+\mathbf{k})
b. Moment about AB. You want the projection of \mathbf{M}_A on AB:
M_{A B}=\mathbf{i} \cdot \mathbf{M}_A=\mathbf{i} \cdot(a P / \sqrt{2})(\mathbf{i}+\mathbf{j}+\mathbf{k})
M_{A B}=a P / \sqrt{2}
You can verify that since AB is parallel to the x axis, M_{A B} is also the x component of the moment M_{A }.
c. Moment about diagonal AG. You obtain the moment of P about AG by projecting M_{A } on AG. If you denote the unit vector along AG by λ (Fig. 2), the calculation looks like this:
\boldsymbol{\lambda}=\frac{\overrightarrow{A G}}{A G}=\frac{a \mathbf{i}-a \mathbf{j}-a \mathbf{k}}{a \sqrt{3}}=(1 / \sqrt{3})(\mathbf{i}-\mathbf{j}-\mathbf{k})
M_{A G}=\boldsymbol{\lambda} \cdot \mathbf{M}_A=(1 / \sqrt{3})(\mathbf{i}-\mathbf{j}-\mathbf{k}) \cdot(a P / \sqrt{2})(\mathbf{i}+\mathbf{j}+\mathbf{k})
M_{A G}=(a P / \sqrt{6})(1-1-1) \quad M_{A G}=-a P / \sqrt{6}
Alternative Method. You can also calculate the moment of P about AG from the determinant form:
M_{A G}=\left|\begin{array}{ccc}\lambda_x & \lambda_y & \lambda_z \\x_{F / A} & y_{F / A} & z_{F / A} \\F_x & F_y & F_z\end{array}\right|=\left|\begin{array}{ccc}1 / \sqrt{3} & -1 / \sqrt{3} & -1 / \sqrt{3} \\a & -a & 0 \\0 & P / \sqrt{2} & -P / \sqrt{2}\end{array}\right|=-a P / \sqrt{6}d. Perpendicular Distance between AG and FC. First note that P is perpendicular to the diagonal AG. You can check this by forming the scalar product P \cdot \lambda and verifying that it is zero:
\mathbf{P} \cdot \boldsymbol{\lambda}=(P / \sqrt{2})(\mathbf{j}-\mathbf{k}) \cdot(1 / \sqrt{3})(\mathbf{i}-\mathbf{j}-\mathbf{k})=(P \sqrt{6})(0-1+1)=0You can then express the moment M_{A G} as -Pd, where d is the perpendicular distance from AG to FC (Fig. 3). (The negative sign is needed because the rotation imparted to the cube by P appears as clockwise to an observer at G.) Using the value found for M_{A G} in part c,
M_{A G}=-P d=-a P / \sqrt{6} \qquad d=a / \sqrt{6}REFLECT and THINK: In a problem like this, it is important to visualize the forces and moments in three dimensions so you can choose the appropriate equations for finding them and also recognize the geometric relationships between them.
