
F


r

r
  rF sin 


Notice that we are multiplying two vectors r and F and the result is a third vector.
Vector Product Cross Product   the product of two vectors that results in a

b
third vector that is perpendicular to the plane
containing the original vectors.
 a measure of the " perpendicularness"
of two vectors
 
a
 
a  b  ab sin 


where  is the angle between a and b
Direction given by
Right Hand Rule – use the fingers of your right hand to rotate the
first vector into the second vector through the
smaller angle between, the vector product is in the
direction of your outstretched thumb
Notice that the vector product is NOT commutative.
Right Hand Rule – use the fingers of your right hand to rotate the
first vector into the second vector through the
smaller angle between, the vector product is in the
direction of your outstretched thumb
 
  r F


A force of F  3.0 N iˆ  4.0 N  ˆj acts on an object located at r  1.0m iˆ  2.0m  ˆj.
What torque does the force provide?
 
  r F
Two Methods
  rF sin 

F

F

Fx
3.0 N   4.0 N 
F  5.0 N
2
2

Fy
1  4.0 N

  tan 

 3.0 N 
  53.1


A force of F  3.0 N iˆ  4.0 N  ˆj acts on an object located at r  1.0m iˆ  2.0m  ˆj.
What torque does the force provide?
Two Methods

rx


r
r
1.0 N   2.0 N 
r  2.24 m
2
2

ry
1  2.0 N


 1.0 N 
  tan 
  63.4


A force of F  3.0 N iˆ  4.0 N  ˆj acts on an object located at r  1.0m iˆ  2.0m  ˆj.
What torque does the force provide?
Two Methods
 
  r F
  rF sin 

F

r
53.1
63.4
116.5
  2.24 m 5.0 N sin 116.5
  10. Nm
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+ counterclockwise torque


A force of F  3.0 N iˆ  4.0 N  ˆj acts on an object located at r  1.0m iˆ  2.0m  ˆj.
What torque does the force provide?
Method 2
 
  r F
  1.0 m  iˆ  2.0 m  ˆj  3.0 N  iˆ  4.0 N  ˆj 
  1.0 m 3.0 N iˆ  iˆ   1.0 m 4.0 N iˆ  ˆj 
  2.0 m 3.0 N  ˆj  iˆ    2.0 m 4.0 N  ˆj  ˆj 
But
  rF sin 
and therefore
iˆ  iˆ  ˆj  ˆj  0?


A force of F  3.0 N iˆ  4.0 N  ˆj acts on an object located at r  1.0m iˆ  2.0m  ˆj.
What torque does the force provide?
Method 2
  1.0 m4.0 N iˆ  ˆj    2.0 m3.0 N  ˆj  iˆ 
  4.0 Nm iˆ  ˆj    6.0 Nm  ˆj  iˆ 
ĵ
k̂
iˆ

iˆ
k̂

ĵ
  4.0 Nm kˆ    6.0 Nm  kˆ 
  10.Nm kˆ