Free Body Diagram

Download Report

Transcript Free Body Diagram

Free Body Diagrams and
Newton’s Laws
Free Body Diagram
A diagram of all the external forces on an object. The
object is drawn as is floating in space (“free”).
•Only include forces on the diagram, not other vectors
such as acceleration or velocity.
•Only include the actual individual forces, not
components or net forces.
Example
Ftable
Free Body Diagram
Fg
Object or System of Objects
System
Problem #1
A 55 N box is at rest on a table. What is the force of the
table on the box?
Ftable
F  0
F  Ftable  Fg
Ftable  Fg  0
Ftable  Fg  55 N
Fg
Problem #2
A 20 kg box is at rest on a table. What is the force of the
table on the box?
Ftable
F  0
F  Ftable  Fg
Ftable  Fg  0
Ftable  Fg  mg
Fg
mg  20  9.8  196 N
Problem #3
Two 20 kg boxes are tied together with a rope. A force of
40 N is applied to the block on the right. What is the
acceleration of the blocks?
Treating the boxes as one system
F  ma
F  40
40   20  20  a
a  1.0 m 2
s
Problem #4
Two 20 kg boxes are tied together with a rope. A force of
40 N is applied to the block on the right. What is the
tension in the rope?
Examine the forces on one box
F  ma
F  40  T
40  T  20 1.0 
T  20 N
Problem #5
Two 20 kg boxes are tied together with a rope. A force of
60 N is applied to the block on the right at an angle of 30
degrees to the horizontal. What is the acceleration of the
blocks?
Fx  ma
Treating the boxes as one system
Fx  60cos30
Fy  0
Fy  Ftable  60sin 30   20  20  9.8
60cos30  40a
51.9  40a
a  1.3 m 2
s
Problem #6
The length of the chain between the car and the tree
is 15.0 m and the perpendicular force causes the
chain to deflect 0.5 m from the dotted line in the
diagram. If the perpendicular force is 100 N what is
the force T on the car if the car doesn’t move?
Examine the point in the chain where the
perpendicular force is applied
 0.5 
o
  sin 
  3.8
 7.5 
1
T
T
Fx  T cos3.8  T cos3.8  0
Fy  T sin 3.8  T sin 3.8  F  0
F┴
2T sin 3.8  100  0
100
T
 754 N
2sin 3.8