Question

A stool at a restaurant is anchored to the floor. When a customer is in the process of sitting down, a horizontal force with magnitude F1 is exerted at the top of the stool support as shown in the figure. (Figure 1) When the customer is seated, a vertical force with magnitude F2 is exerted on the stool support. If the maximum moment magnitude that the stool support can sustain about point A is MA = 180lb?ft , what is the maximum height d1 that the stool can have if the magnitudes of the two forces are F1 = 70.0lb andF2 = 180lb ? Assume that moments acting counterclockwise about point A are positive whereas moments acting clockwise about A are negative. Express your answer numerically in feet to three significant figures. To understand the concept of moment of a force and how to calculate it using a scalar formulation. The magnitude of the moment of a force with a magnitude F around a point O is defined as follows: where d is the force's moment arm. The moment arm is the perpendicular distance from the axis at point O to the force's line of action

A stool at a restaurant is anchored to the floor. When a customer is in the process of sitting down, a horizontal force with magnitude F1 is exerted at the top of the stool support as shown in the figure. (Figure 1) When the customer is seated, a vertical force with magnitude F2 is exerted on the stool support. If the maximum moment magnitude that the stool support can sustain about point A is MA = 180lb?ft , what is the maximum height d1 that the stool can have if the magnitudes of the two forces are F1 = 70.0lb andF2 = 180lb ? Assume that moments acting counterclockwise about point A are positive whereas moments acting clockwise about A are negative.

Express your answer numerically in feet to three significant figures.

To understand the concept of moment of a force and how to calculate it using a scalar formulation.

The magnitude of the moment of a force with a magnitude F around a point O is defined as follows:

MO=Fd

where d is the force's moment arm. The moment arm is the perpendicular distance from the axis at point O to the force's line of action.

Answer 1

Concepts and reason

Magnitude of moment can be calculated by multiplying the magnitude of force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.

Moment of a force about a point is given by cross product of position vector of any point on the line of action of force from the point about which moment has to be calculated and force vector.

Vector form of representation of location of a point from a reference point is called position vector.

A body is in equilibrium if vector sum of all the forces is equal to zero or moment of all force vectors about any point is equal to zero.

Fundamentals

Procedure to write position vector of point, $A = \left( {{x_A},\;{y_A}\;,{z_A}\;} \right)$ with reference to origin is as follows:

${{\bf{r}}_{{\bf{OA}}}} = \left( {{x_A}{\bf{i}} + {y_A}{\bf{j}} + {z_A}{\bf{k}}} \right)$

Here, coordinates of point A along x, y, and z axis are ${x_A},\;{y_A}\;,{\rm{ and }}{z_A}$ respectively and unit vectors along x, y, and z axis are ${\bf{i}}\;,{\bf{j}},\;{\rm{and}}\;{\bf{k}}$ respectively.

Procedure to write position vector of point, $A = \left( {{x_A},\;{y_A}\;,{z_A}\;} \right)$ with reference to point $B = \left( {{x_B},\;{y_B}\;,{z_B}\;} \right)$ is as follows:

$\begin{array}{c}\\{{\bf{r}}_{{\bf{AB}}}} = {{\bf{r}}_{{\bf{OA}}}} - {{\bf{r}}_{{\bf{OB}}}}\\\\{{\bf{r}}_{{\bf{AB}}}} = \left( {{x_A}{\bf{i}} + {y_A}{\bf{j}} + {z_A}{\bf{k}}} \right) - \left( {{x_B}{\bf{i}} + {y_B}{\bf{j}} + {z_B}{\bf{k}}} \right)\\\\ = \left( {{x_A} - {x_B}} \right){\bf{i}} + \left( {{y_A} - {y_B}} \right){\bf{j}} + \left( {{z_A} - {z_B}} \right){\bf{k}}\\\end{array}$

Here, coordinates of point B along x, y, and z axis are ${x_B},\;{y_B}\;,{\rm{ and }}{z_B}$ respectively

Equation for calculation of magnitude of vector, ${\bf{r}} = \left( {{r_x}{\bf{i}} + {r_y}{\bf{j}} + {r_z}{\bf{k}}} \right)$ is as follows:

$\left| {\bf{r}} \right| = \sqrt {{{\left( {{r_x}} \right)}^2} + {{\left( {{r_y}} \right)}^2} + {{\left( {{r_z}} \right)}^2}}$

Here, ${r_x}$ is the x-component of the position vector (coefficient of i component of position vector), ${r_y}$ is the y-component of the position vector (coefficient of j component of position vector), and ${r_z}$ is the z-component of the position vector (coefficient of k component of position vector).

Calculation of unit vector for the direction along which the force applied is prominent.

Equation used for calculation of unit vector.

${\bf{u}} = \frac{{\bf{r}}}{{\left| {\bf{r}} \right|}}$

Moment about a point can be calculated in two ways.

First way:

It can be calculated by multiplying the magnitude of force with the distance between the point of application of force and point about which moment is to be calculated.

Second way:

Use vector notation of distance and force. The cross product of this force vector and distance vector would give the moment in Cartesian vector form.

Basic equations for the calculation of internal forces are as follows:

Equation for calculation of moment about point B due force applied at point F.

${\vec M_A} = {\vec r_{AC}} \times \vec F$

Here, ${\vec r_{AC}}$ is the position of C with respect to A, ${\vec M_A}$ is the moment, and $\vec F$ is the force.

General rule for cross product of the vectors is as follows:

$\begin{array}{l}\\\vec i \times \vec i = 0\\\\\vec i \times \vec j = \vec k\\\\\vec i \times \vec k = - \vec j\\\\\vec j \times \vec j = 0\\\\\vec j \times \vec i = - \vec k\\\\\vec j \times \vec k = \vec i\\\\\vec k \times \vec k = 0\\\\\vec k \times \vec i = \vec j\\\\\vec k \times \vec j = - \vec i\\\end{array}$

General sign convection for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

General sign convection for axis: Distance along the axis is positive and opposite to the axis is negative.

Draw the free body diagram of the stool.

Calculate the moment for the force about point A.

$\begin{array}{l}\\\sum {{M_A} = 0} \\\\{M_A} = {F_1}\left( {{d_1}} \right) + {F_2}\left( 0 \right)\\\\180 = 70\left( {{d_1}} \right) + 180\left( 0 \right)\\\\180 = 70\left( {{d_1}} \right)\\\\{d_1} = 2.571\,{\rm{ft}}\\\end{array}$

Ans:

Therefore, the perpendicular distance from the axis at point O to the forces line of action is $2.571\,{\rm{ft}}$ .