Moment of a force along an axis Couples “If you find yourself in a hole, stop digging.” –Will Rogers Objectives ¢ Understand the vector formulation for finding the component of a moment along an axis ¢ Understand the idea of a couple and the moment it produces 2 Moments Along an Axis, Couples Monday, September 24, 2012 1 Tools ¢ Basic Trigonometry ¢ Pythagorean Theorem ¢ Algebra ¢ Visualization ¢ Position Vectors ¢ Unit Vectors ¢ Reviews ¢ Cross Products ¢ Dot Products 3 Moments Along an Axis, Couples Monday, September 24, 2012 Review ¢ A moment is the tendency of a force to cause rotation about a point or an axis 4 Moments Along an Axis, Couples Monday, September 24, 2012 2 Moment about an Axis ¢ There are times that we are interested in the moment of a force that produces some component of rotation about (or along) a specific axis ¢ We can use all the we have learned up to this point to solve this type of problem 5 Moments Along an Axis, Couples Monday, September 24, 2012 Moment about an Axis ¢ First select any point on the axis of interest and find the moment of the force about that point ¢ Using the dot product and multiplication of the scalar times the unit vector of the axis, the component of the moment about the axis can be calculated 6 Moments Along an Axis, Couples Monday, September 24, 2012 3 Moment about an Axis ¢ If we have an axis a-a we can find the component of a moment along that axis by M = u u M a−a a−a ( a−a ) where M is the moment about any point on a-a 7 Moments Along an Axis, Couples Monday, September 24, 2012 Problem F4-17 8 Moments Along an Axis, Couples Monday, September 24, 2012 4 Example ¢ Given: System as Shown 9 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Required: Moment of force F along axis a-a 10 Moments Along an Axis, Couples Monday, September 24, 2012 5 Example ¢ Solution: M = u u M a− a a− a ( a− a ) 11 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Solution: M = u u M a−a a−a ( a−a ) need u and M a−a where M is the moment of the force about some point on a-a 12 Moments Along an Axis, Couples Monday, September 24, 2012 6 Example ¢ Solution: r u = 0 A a−a r 0 A r A O rOA 0 A = { } − { } {A} = {−3,−2,6} ft {O} = {0,0,0} ft r = −3i − 2 j + 6k ft 0 A ( ) r0 A = 7 ft 3 2 6 u = − i − j + k a−a 7 7 7 13 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Solution: Now we have to select some point on a-a and find the moment of the force F about that point ¢ Please don’t mistake rOA for the moment arm r ¢ rOA is just the position OA vector that we used to generate the unit vector along a-a 14 Moments Along an Axis, Couples Monday, September 24, 2012 7 Example ¢ The origin is on a-a (not always but in this problem it is) so we can find the moment of F about the origin rOA 15 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ First we define a moment arm from the origin to the line of action of F rOA 16 Moments Along an Axis, Couples Monday, September 24, 2012 8 Example ¢ We can use rOB as our moment arm r = B − O OB { } { } rOA {B} = {0,4,0} ft r B {O} = {0,0,0} ft OB r = 4 j ft OB ( ) C 17 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Now we have to describe F as a Cartesian vector F = Fu F u = u F CB r u = CB CB r rOA CB r B C CB = { } − { } {B} = {0,4,0} ft {C} = {4,0,−2} ft r B OB r = −4i + 4 j + 2k ft CB ( ) rCB = 6 ft 4 4 2 u = − i + j + k C F 6 6 6 F = −400i + 400 j + 200k lb ( ) 18 Moments Along an Axis, Couples Monday, September 24, 2012 9 Example ¢ Taking the moment of F about the origin (O) M = rOB ⊗ F rOA M = (4 j) ft ⊗(−400i + 400 j + 200k)lb M = ((4 j ⊗ −400i) + (4 j ⊗ 400 j) + (4 j ⊗ 200k)) ft ⋅lb M = 1600k + 800i ft ⋅lb B ( ) rOB C 19 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Finding the magnitude of the projection of M along a-a ua−a M 3 2 6 u = − i − j + k r a−a 7 7 7 OA M = (1600k + 800i) ft ⋅lb ⎛ 3⎞ ⎛ 6⎞ r B u M = − 800 ft ⋅lb + 1600 ft ⋅lb OB a−a ⎝⎜ 7⎠⎟ ( ) ⎝⎜ 7⎠⎟ u M = 1028.57 ft ⋅lb a−a C 20 Moments Along an Axis, Couples Monday, September 24, 2012 10 Example ¢ Converting the magnitude to a Cartesian vector u u M a−a ( a−a ) 3 2 6 u = − i − j + k a−a 7 7 7 rOA ua−a M = 1028.57 ft ⋅lb u u M = −440.82i − 293.88 j + 881.63k ft ⋅lb a−a ( a−a ) ( ) B rOB C 21 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ To find the magnitude of the projection, we can also use what is known as the mixed product rOA B rOB C 22 Moments Along an Axis, Couples Monday, September 24, 2012 11 Example ¢ Again, starting with the force F the moment arm rOB and the unit vector along the axis of concern uOA rOA B rOB C 23 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Taking advantage of the properties of the dot product and the cross product, we can set up the problem in a matrix form as uuu xyz rOA rrr r B xyz OB FFFxyz C 24 Moments Along an Axis, Couples Monday, September 24, 2012 12 Example ¢ Notice that since there are no vectors in this expression, the result of reducing the matric will be a scalar. uuu xyz rOA rrr r B xyz OB FFFxyz C 25 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ We can use the same method that we used when we calculated the cross product ¢ Here we use it to find the magnitude of the moment along the axis uuu xyz rOA rrr r B xyz OB FFFxyz C 26 Moments Along an Axis, Couples Monday, September 24, 2012 13 Example ¢ Substituting the coefficients of the unit vector 326 −− 777 rOA rrrxyz B rOB FFFxyz C 27 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Substituting the coefficients of the moment arm 3 2 6 − − 7 7 7 rOA 0 ft 4 ft 0 ft B rOB F F F x y z C 28 Moments Along an Axis, Couples Monday, September 24, 2012 14 Example ¢ Substituting the coefficients of the force 3 2 6 − − 7 7 7 rOA 0 ft 4 ft 0 ft r B −400lb 400lb 200lb OB C 29 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Using our method for reducing the matrix 3 2 6 3 2 − − − − 7 7 7 7 7 0 ft 4 ft 0 ft 0 ft 4 ft rOA −400lb 400lb 200lb −400lb 400lb B rOB C 30 Moments Along an Axis, Couples Monday, September 24, 2012 15 Example ¢ Using our method for reducing the matrix 3 2 6 3 2 − − − − 7 7 7 7 7 0 ft 4 ft 0 ft 0 ft 4 ft −400lb 400lb 200lb −400lb 400lb r 3 OA − 4 ft 200lb − 400lb 0 ft 7 (( )( ) ( )( )) 2 r B − 0 ft −400lb − 200lb 0 ft OB 7 (( )( ) ( )( )) 6 + ((0 ft)(400lb) − (−400lb)(4 ft)) C 7 31 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Using our method for reducing the matrix 3 2 6 3 2 − − − − 7 7 7 7 7 0 ft 4 ft 0 ft 0 ft 4 ft rOA −400lb 400lb 200lb −400lb 400lb 3 2 6 r B − (800 ft • lb) − (0) + (1600 ft • lb) OB 7 7 7 C 32 Moments Along an Axis, Couples Monday, September 24, 2012 16 Example ¢ The magnitude of the moment produced by F along a-a is then 3 2 6 3 2 − − − − 7 7 7 7 7 0 ft 4 ft 0 ft 0 ft 4 ft r −400lb 400lb 200lb −400lb 400lb OA 3 2 6 − 800 ft • lb − 0 + 1600 ft • lb ( ) ( ) ( ) B 7 7 7 rOB 7200 ft • lb = = 1028.57 ft • lb 7 C 33 Moments Along an Axis, Couples Monday, September 24, 2012 Example ¢ Take care in identifying the unit vector of the force and the position vector that is the moment arm 3 2 6 3 2 − − − − 7 7 7 7 7 r 0 ft 4 ft 0 ft 0 ft 4 ft OA −400lb 400lb 200lb −400lb 400lb 3 2 6 r B − 800 ft • lb − 0 + 1600 ft • lb OB 7 ( ) 7 ( ) 7 ( ) 7200 ft • lb = = 1028.57 ft • lb C 7 34 Moments Along an Axis, Couples Monday, September 24, 2012 17 Example ¢ Remember that the unit vector will not have units and that the position vector will have units 3 2 6 3 2 − − − − 7 7 7 7 7 r 0 ft 4 ft 0 ft 0 ft 4 ft OA −400lb 400lb 200lb −400lb 400lb 3 2 6 r B − 800 ft • lb − 0 + 1600 ft • lb OB 7 ( ) 7 ( ) 7 ( ) 7200 ft • lb = = 1028.57 ft • lb C 7 35 Moments Along an Axis, Couples Monday, September 24, 2012 Couples ¢ A couple is a system of two forces ¢ The forces must satisfy the following conditions for the force system to be a couple l The magnitudes of the forces must be the same l The unit vectors of the forces must be opposite in direction u1 1 u2 FF12= = (− ) 36 Moments Along an Axis, Couples Monday, September 24, 2012 18 Couples ¢ A couple produces pure rotation on a system ¢ The sum of the two forces will be equal to 0 u1 1 u2 FF12= = (− ) 37 Moments Along an Axis, Couples Monday, September 24, 2012 Couples ¢ Most importantly, a couple produces the same amount of rotation about any moment center u1 1 u2 FF12= = (− ) 38 Moments Along an Axis, Couples Monday, September 24, 2012 19 Couples ¢ We will start with two forces in space who satisfy the conditions of a couple FF12= F u F 1 u F 1 1 = (− ) 2 F2 39 Moments Along an Axis, Couples Monday, September 24, 2012 Couples ¢ We will then pick some point in space as the moment center FF12= a F1 u F 1 u F 1 = (− ) 2 F2 40 Moments Along an Axis, Couples Monday, September 24, 2012 20 Couples ¢ We can pick a point on each force’s line of action and construct moment arms from point a to a point on the life of action on each of these forces r b a ab F1 FF12= r ac F2 u F = −1 u F 1 ( ) 2 c 41 Moments Along an Axis, Couples Monday, September 24, 2012 Couples ¢ The total moment generated by the two forces about a is equal to M a = rab ⊗ F1 + rac ⊗ F2 r b a ab F1 rac F2 c u F 1 u F 1 = (− ) 2 FF12= 42 Moments Along an Axis, Couples Monday, September 24, 2012 21 Couples ¢ Substituting the definition of F1 and F2 M = r ⊗ u F + r ⊗ u F a ab ( 1 1 ) ac ( 2 2 ) r b a ab F1 rac F2 c u F 1 u F 1 = (− ) 2 FF12= 43