4: 4-59

Problem: Determine the magnitude of force F in cable AB in order to produce a moment of 500 lb * ft about the hinged axis CD, which is needed to hold the panel in the position shown.

Variable Solution:

Givens: 
a = 4 [ft]
b = 6 [ft]
c = 6 [ft]
M_CD = 500 [lb⋅ft]

Numerical Solution:
F = M_CD⋅(4a²+b²)^1/2⋅(csin(tan^-1((4b²+a²)^1/2/c)))^-1⋅(2acos(tan^-1(a/(2b)))-bsin(tan^-1(a/(2b))))^-1
→(500 [lb⋅ft])⋅(4(4 [ft])²+(6 [ft])²)^1/2⋅((6 [ft])sin(tan^-1((4(6 [ft])²+(4 [ft])²)^1/2/(6 [ft]))))^-1⋅(2(4 [ft])cos(tan^-1((4 [ft])/(2(6 [ft]))))-(6 [ft])sin(tan^-1((4 [ft])/(2(6 [ft])))))^-1
F = 162.0 [lb]

4: 4-47

Problem: Determine the magnitude of the moment of each of the three forces about the axis AB. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach

Variable Solution:

Givens: 
F1 = 60 [N]
F2 = 85 [N]
F3 = 45 [N] 
a = 1.5 [m]
b = 2 [m]

Numerical Solution:
M_AB(F1) (scalar) = b⋅F1⋅sin(tan^-1(a/b))
→(2 [m])(60 [N])⋅sin(tan^-1((1.5 [m])/(2 [m])))
M_AB(F1) = 72 [N⋅m] 

M_AB(F1) (vector) = ab⋅f1⋅(a²+b²)^-1/2
→(1.5 [m])(2 [m])(60 [N])⋅((1.5 [m])²+(2 [m])²)^-1/2
M_AB(F1) = 72 [N⋅m] 

M_AB(F2) (scalar) = 0
M_AB(F2) (vector) = 0

M_AB(F3) (scalar) = 0
M_AB(F3) (vector) = 0

4: 4-21

Problem: In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 500 lb * in about point A. Determine the required magnitude of force F.

Variable Solution:

Givens: 
a = 18 [in]
b = 5 [in]
α = 30
M_A = 500 [lb⋅in] CW
**notice that M_A is clockwise so it will be negative in our numerical solution, because it goes into the page

Numerical Solution:
F = -M_A/(bsinα+acosα)
→-(-500 [lb⋅in]) / ((5 [in])sin(30)+(18 [in])cos(30))
F = 27.64 [lb]

4: 4-5

Problem: Determine the moment about point B of each of the three forces acting on the beam

Variable Solution:

Givens: 

F1 = 375 [lb]

F2 = 500 [lb]

F3 = 160 [lb]

a = 8 [ft]

b = 6 [ft]

c = 5 [ft]

d = 0.5 [ft] 

α = 30

Numerical Solution:

M_BF1 = F1(b+c) → (375 [lb])(6 [ft] + 5 [ft]) 

M_BF1 = 4125 [lb⋅ft] CCW

M_BF2 = cF2cos(tan^-1(3/4)) → (5 [ft])(500 [lb])cos(tan^-1(3/4)) 

M_BF2 = 2000 [lb⋅ft] CCW

M_BF3 = dF3sinα → (0.5 [ft])(160 [lb])sin(30) 

M_BF3 = 40 [lb⋅ft] CCW

4: F4-2

Problem: Determine the moment of the force about point O

4: F4-1

Problem: Determine the moment of the force about point O

notice the scalar solution was a scalar value and the vector solution was a vector. 

The vector is negative because it actually goes into the page, but taking the magnitude of it would result in the same magnitude as the scalar solution. 

4: Notes

Force System Resultants​

1) Moment of Force ​

A force produces a turning effect or moment about a point O that does not lie on its line of action
The direction of the moment is defined using the right hand rule. M_O always acts along an axis perpendicular to the plane containing F & d, and passes through the point O. 

The easiest way I remember the right hand rule is by the bottle cap method that my professor mentioned to me. If you're holding a soda bottle in your left hand and unscrewing it with your right, the cap spins from left to right and moves in an upward direction. 

a) Scalar Definition​

The magnitude of the moment of force is the product of the force and the moment arm, or perpendicular distance from point O to the line of action of the force. 
Rather than finding d, it is normally easier to resolve the force into its x and y components, determine the moment of each component about the point, and then sum the results. This is called the principle of moments.

M_O = Fd = F_xy - F_yx

b) Vector Definition​

Most common with 3D moment analysis. 
The moment is determined by taking the cross product of the vectors. 
if r is a position vector extending from point O to any point A, B, or C on the line of action of a force 
F, 

M_O = r_A × F = r_B × F = r_C × F ​

Cross Product: 
if U = <a, b, c> & V = <d, e, f>, then U × V = <bf - ce, cd - af, ae - bd>

2) Moment about an Axis​

If the moment of a force F is to be determined about an arbitrary axis a, then for a scalar solution the moment arm, or shortest distance d_a from the line of action of the force to the axis must be used. This distance is perpendicular to both the axis and the force line of action. 

In 3D, the scalar triple product should be used. It's best to use a unit vector, u_a, that specifies the direction of the axis and a position vector, r, that is directed from any point on the axis to any point on the line of action of the force. 

If M_a is calculated as a negative scalar, then the sense of direction of M_a is opposite to u_a

M_a = Fd_a = u_a⋅(r × F)

3) Couple Moment​

Consists of two equal but opposite forces that act at a perpendicular distance d apart. Couples tend to produce a rotation without translation. 

The magnitude of the couple moment is M = Fd, and its direction is established using the right hand rule. 

If the vector cross product is used to determine the moment of a couple, then r extends from any point on the line of action of one of the forces to any point on the line of action of the other force F that is used in the cross product. 
M = r × F

4) Simplification of a Force and Couple System​

Any system of forces and couples can be reduced to a single resultant force and resultant couple moment acting at a point. The resultant force is the sum of all the forces in the system, F_R = ΣF, and the resultant couple moment is equal to the sum of all the moments of the forces about the point and couple moments.

M_RO = ΣM_O + ΣM​

Equate the moment of the resultant force about the point to the moment of the forces and couples in the system about the same point

If the resultant force and couple moment at a point are not perpendicular to one another, then this system can be reduced to a wrench, which consists of the resultant force and collinear couple moment. 

5) Coplanar Distributed Loading​

A simple distributed loading can be represented by its resultant force, which is equivalent to the area under the loading curve. This resultant has a line of action that passes through the centroid or geometric center of the area or volume under the loading diagram.

3: Notes

Particle Equilibrium​

For a particle to be in equilibrium, all forces acting on the particle form a zero resultant force
It is necessary to draw a free body diagram to account for all the forces that act on a particle
F_R = ΣF = 0

2D​

The two scalar equations of force equilibrium can be applied with reference to an established x, y coordinate system

ΣF_x = 0
ΣF_y = 0​

TENSION & pulleys - The tension force in a continuous cable that passes over a pulley is constant throughout the cable to keep it in equilibrium
SPRING - If the problem involves an elastic spring, then the stretch or compression (s) of the spring can be related to the force applied to it

F = ks​

3D​

First express each force on the free-body diagram as a Cartesian vector, then when the forces are summed and set equal to zero, the i j & k components are also zero. 

ΣF_NET = 0
ΣF_x = 0
ΣF_y = 0
ΣF_z = 0
​

2: Notes

Scalars & Vectors​

Scalar - any positive or negative physical quantity that can be completely specified by its magnitude 
Vector - any physical quantity that requires both a magnitude and a direction for its complete description

The resultant of several coplanar forces can easily be determined if an x, y coordinate system is established and the forces are resolved along the axes
Multiplication or division of a vector by a scalar will change only the magnitude of the vector. 
If vectors are collinear, the resultant is simply the algebraic or scalar addition

Parallelogram Law​

Two forces add according to the parallelogram law. The components form the sides of the parallelogram and the resultant is the diagonal. 

Cartesian Vectors​

Representation​

A vector V is represented by its rectangular components in each of the xyz axes in the form 

V=V_xi+V_yj+V_zk​

Magnitude​

|V|=(V_x²+V_y²+V_z²)^½

Direction​

the direction of V is defined by the coordinate direction angles α, β, and γ measured between the tail of V and the positive x, y, and z axes, respectively

cosα = V_x/V
cosβ = V_y/V
cosγ = V_z/V​

an easy way of obtaining these direction cosines is to form a unit vector U_V in the direction of V

U_V = V/|V| = V_x/V i + V_y/V j + V_z/V k​

an important relation among the direction cosines can be formulated as 

cos²α + cos²β + cos²γ = 1​

Addition​

the addition or subtraction of two or more vectors is greatly simplified if the vectors are expressed in terms of their Cartesian components. A resultant vector R from the addition of two vectors is

V_R = ΣV = ΣV_xi + ΣV_yj + ΣV_zk​

Position Vectors​

a position vector r is defined as a fixed vector which locates a point in space relative to another point
if r extends from the origin of coordinates, O, to point P(x,y,z)

r=xi+yj+zk​

for two vectors r_A and r_B, the position vector represented by the two of them is
r=(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

The easiest way to formulate the components of a position vector is to determine the distance and direction that must be traveled along the x,y,z directions - going from the tail to the head of the vector

A force F acting in the direction of a position vector r can be represented in Cartesian form if the unit vector U of the position vector is determined and it is multiplied by the magnitude of the force, i.e. F=|F|u=|F|(r/|r|)

Dot Product​

The dot product is used to determine the angle between two vectors or the projection of a vector in a specified direction

A⋅B = ABcosθ = A_xB_x + A_yB_y + A_zB_z​

The magnitude of the projection of vector A along a line a whose direction is specified by u_a is determined from the dot product |A_a| = A⋅u_a