Question

Using the principle of transmissibility, the force F can be slid along its line of action to point B. At point B, the force F creates an angle φ with the shed. (Part C figure) What is Mfx , the contribution to the moment about point O made by the x component of the force at point B? What is Mfy , the contribution to the moment about point O made by the y component of the force Fat point B? What is the total moment Mdue to the force F about point O? Assume that moments acting counterclockwise about point O are positive whereas moments acting clockwise are negative.

As shown, a rope is attached to a l=19.0 ft high shed that is to be relocated. A man pulls on the end of the rope at point A; the rope is attached to the shed at point B. (Intro 1 figure) As the man pulls on the rope, it creates an angle theta with the horizontal. The end of the rope is located at x=10.0 ft from the shed and y=4.50 ft off the ground. The man pulls on the rope with a force of magnitude F= 70 lbs.

Answer 1

Concepts and reason

Free body diagram:

These diagrams are used to show all the external forces which are applied on the body. It is basically a vector representation of all the external forces. The dynamic analysis is done based on the free body diagram of any physical object.

Moment of a force:

Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The moment’s magnitude can be determined from the product of force’s magnitude and the distance to the force measured perpendicularly.

The magnitude of the developed moment will be zero, if the line of action of the applied force is along the direction of the point about which the moment is to be determined.

The angle of inclination by the force $F$ can be determined using the inverse tangent relation. Then, resolve the force F into x and y components. Thus, calculate the moment at point O about x and y directions as the product of force and perpendicular distances using principle of transmissibility.

Fundamentals

Resolution of force vector:

Here, the angle of inclination or the direction of force with horizontal is $\theta$ .

For a force vector ${\bf{F}}$ acting as shown in the Figure (1), the components of force along the horizontal x and vertical y directions is given by,

$\begin{array}{l}\\{F_x} = F\cos \theta \\\\{F_y} = F\sin \theta \\\end{array}$

In a right triangle, write the basic trigonometric relation for an inclination as follows:

$\tan \theta = \frac{{{\rm{opposite side}}}}{{{\rm{adjacent side}}}}$

The equation for the moment $\left( M \right)$ about any point is given as follows:

$M = Fd$

Here, the applied force is $F$ and the perpendicular distance between the force and reference point is $d$ .

The resultant moment $\left( {{M_R}} \right)$ for two mutually perpendicular moment components is given as follows:

${M_R} = \sqrt {{M_x}^2 + {M_y}^2}$

Here, the moment component about $x{\rm{ - }}$ axis is ${M_x}$ and the moment component about $y{\rm{ - }}$ axis is ${M_y}$ .

General sign conventions:

Moment in clockwise direction is considered as negative and moment in counter clockwise direction is considered as positive.

Draw the free body diagram for the force action as shown in Figure (2).

Here, the horizontal distance of the man from point O is $x$ , the vertical distance of the man from point O is $y$ and the angle made by the force $F$ with the vertical axis is $\phi$ .

Calculate the angle of inclination by the force $F$ with vertical axis.

$\tan \phi = \frac{x}{{l - y}}$

Substitute $10.0\;{\rm{ft}}$ for $x$ , $19.0\;{\rm{ft}}$ for $l$ and $4.50\;{\rm{ft}}$ for $y$ .

$\begin{array}{l}\\\tan \phi = \frac{{10.0\;{\rm{ft}}}}{{19.0\;{\rm{ft}} - 4.50\;{\rm{ft}}}}\\\\\phi = {\tan ^{ - 1}}\left( {0.68956} \right)\\\\\phi = 34.592^\circ \\\end{array}$

Calculate the $x{\rm{ - }}$ component of force at point B.

${F_x} = F\sin \phi$

Here, the $x{\rm{ - }}$ component of force at point B is ${F_x}$ .

Determine the contribution of the moment about point O made by the $x{\rm{ - }}$ component of the force at point B $\left( {{M_{fx}}} \right)$ .

${M_{fx}} = {F_x} \cdot l$

Substitute $F\sin \phi$ for ${F_x}$ .

${M_{fx}} = \left( {F\sin \phi } \right) \cdot l$

Substitute $70\;{\rm{lbs}}$ for $F$ , $34.592^\circ$ for $\phi$ and $19.0\;{\rm{ft}}$ for $l$ .

$\begin{array}{l}\\{M_{fx}} = \left( {70\;{\rm{lbs}}} \right) \times \left( {\sin 34.592^\circ } \right) \times \left( {19.0\;{\rm{ft}}} \right)\\\\{M_{fx}} = 755.079\;{\rm{lb - ft}}\\\end{array}$

Calculate the $y{\rm{ - }}$ component of force at point B.

${F_y} = F\cos \phi$

Here, the $y{\rm{ - }}$ component of force at point B is ${F_y}$ .

Determine the contribution of the moment about point O $\left( {{M_{fy}}} \right)$ made by the $y{\rm{ - }}$ component of the force at point B.

${M_{fy}} = {F_y} \cdot d$

Here, the perpendicular distance of the point of application of force from point O is $d$ .

Substitute $F\cos \phi$ for ${F_y}$ .

${M_{fy}} = \left( {F\cos \phi } \right) \cdot d$

Substitute $70\;{\rm{lbs}}$ for $F$ , $34.592^\circ$ for $\phi$ and $0$ for $d$ .

$\begin{array}{l}\\{M_{fy}} = \left( {70\;{\rm{lbs}}} \right) \times \left( {\cos 34.592^\circ } \right) \times \left( 0 \right)\\\\{M_{fy}} = 0\\\end{array}$

Calculate the total moment $\left( {{M_o}} \right)$ due to the force $F$ about point O.

${M_o} = \sqrt {{M_{fx}}^2 + {M_{fy}}^2}$

Substitute $755.079\;{\rm{lb - ft}}$ for ${M_{fx}}$ and $0$ for ${M_{fy}}$ .

$\begin{array}{l}\\{M_o} = \sqrt {{{\left( {755.079\;{\rm{lb - ft}}} \right)}^2} + {{\left( {0\;{\rm{lb - ft}}} \right)}^2}} \\\\{M_o} = 755.079\;{\rm{lb - ft}}\\\end{array}$

Ans:

The contribution to the moment about point O $\left( {{M_{fx}}} \right)$ made by the $x{\rm{ - }}$ component of the force at point B is $755.079\;{\rm{lb}} \cdot {\rm{ft}}$ .

The contribution to the moment about point O $\left( {{M_{fy}}} \right)$ made by the $y{\rm{ - }}$ component of the force at point B is 0.

The total moment $\left( {{M_o}} \right)$ due to the force $F$ about point O is $755.079\;{\rm{lb}} \cdot {\rm{ft}}$ .

Answer 2

Fx = Fsin(phi)
Fy = Fcos(phi)

MFx = Fx X l = Flsin(phi)
MFy = 0 since distance and force are in same direction

Total moment = Flsin(phi)