3_3_f

Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Take F1 = {400i + 300j + 120 k} N.

3_3_e

The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A.

3_3_d

Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support of C. This requires a moment of M = 80 lbft to be developed at C.

3_3_c

The pole supports a 22-lb traffic light. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A.

3_3_b

The curved rod lies in the x-y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point B.

3_3_a

The curved rod lies in the x-y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point O.

3_2_h

If it takes a force of F = 125 lb to pull the nail out, determine the smallest vertical force P that must be applied to the handle of the crowbar. Hint: this requires the moment of F about point A to be equal to the moment of P about A. Why?

3_2_g

 If the resultant moment about point A is 4800 N m clockwise, determine the magnitude of F3 if F1 = 300 N and F2 = 400 N

3_2_f

The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force of 50 N is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that is creates the opposite moment about C?

3_2_e

The Snorkel Co. produces the articulating boom platform that can support a weight of 550 lb. If the boom is in the position shown, determine the moment of this force about points A, B, and C.

3_2_d

A force of 40 N is applied to the wrench. Determine the moment of this force about point O. Solve the problem using both scalar analysis and a vector analysis.

3_2_c

If the force F = 100 N, determine the angle  (090o) so that the force develops a clockwise moment about point O of 20 Nm.

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.