Self Check: Statics
First:
If you have to:
As a last resort:
Try with no notes or neighbors.
Use your notes and/or textbook.
Ask your neighbors for help.
The purpose of this is for YOU to figure out WHAT YOU DON’T KNOW!
‘Cause you will need to know how to do these on the test!
SC 1: Forces in equilibrium.
Forces 1, 2, and 3 are in equilibrium. Force 1 has a magnitude of 35 N
and is acting at an angle of 57° above the positive x-axis. Force 2 is 25 N
along the negative y-axis.
a.
b.
c.
d.
e.
f.
Define equilibrium (WRT forces).
Sketch the problem (assumed that you will do this)
Label the diagram (again, assumed!)
Determine the sum of forces 1 and 2.
Find force 3.
If the angle of force 1 is decreased while force 2 remains unchanged, how will
force 3 initially react? Be specific, describing the effect on angle and magnitude
of both x and y components.
g. If the magnitude of force 2 increases, how will force 3 initially react? Again, be
specific.
Key:
a. Forces in balance.
Forces that sum to zero.
F1: 35N at 57°
F3: ???
b.
c.
d.
e.
f.
F2: 25N on -y
Labels above.
x  axis :
See box:
F1x  F1 cos(57  )  19.0623N
See box:
The angle will
F2 x  0
decrease
Fx , NET  F1x  F2 x  19.1N
[move toward the
x-axis and
y  axis :
eventually move

above the x-axis]. F1 y  F1 sin( 57 )  29.3534 N
The magnitude in F2 y  25 N
the x direction will Fy , NET  F1x  F2 x  4.35346 N  4.35 N
increase.
The magnitude in
RESULTANT ( Part d ) :
the y direction will
magnitude  19.12  4.35 2  19.5531  19.6 N
decrease
[move toward the
 4.35 

angle  tan 1 
  12.8646  12.9 above positive x  axis
x-axis and
 19.1 
eventually move
FORCE 3 :
above the x-axis]. 19.6 N at 12.9 below negative x  axis
g. The angle will
decrease.
[move toward the x-axis and eventually move above the x-axis].
The magnitude in the x direction will remain unchanged.
The magnitude in the y direction will decrease.
[move toward the x-axis and eventually move above the x-axis].
First:
If you have to:
As a last resort:
Try with no notes or neighbors.
Use your notes and/or textbook.
Ask your neighbors for help.
The purpose of this is for YOU to figure out WHAT YOU DON’T KNOW!
‘Cause you will need to know how to do these on the test!
SC 2: Torques
A 65 kg student exerts a force of 55 N on the end of a 25 kg door 74 cm wide. What is
the magnitude of the torque if the force is exerted (a) perpendicular to the door, and (b)
at a 30º angle to the face of the door? Is the door in equilibrium? Explain!
Now the student’s older sibling (who didn’t take physics!) pushes against the door to
resist the motion. She doesn’t understand torques, and therefore pushes against the door
in its center not at its end. At least she pushes perpendiculary to the door! With what
force must she push to balance the force applied by her younger sibling in both
problems above? Is the door in equilibrium? Explain!
KEY:
a.   F d  55N 0.74m  40.70  40.7 Nm
b.   F d  55N sin( 30 )0.74m  20.35  20.4 Nm
c. The door is not in equilibrium as there is no counter (or “anti”?) torque.
One torque cannot be in equilibrium!
  0
d.
1   2  0
55 N 0.74m  F2 (0.74m / 2)  0
F2  110 N
  0
e.
1   2  0
55N sin( 30 )0.74m  F

F2  55 N
2
(0.74m / 2)  0
First:
If you have to:
As a last resort:
Try with no notes or neighbors.
Use your notes and/or textbook.
Ask your neighbors for help.
The purpose of this is for YOU to figure out WHAT YOU DON’T KNOW!
‘Cause you will need to know how to do these on the test!
SC 3: Forces and Torques in equilibrium: Using both simultaneously!
[P(9):20] A shop sign weighing 245 N is supported by a uniform 155-N beam as shown
in Fig. 9–54. Determine the:
a. mass of the beam [15.8],
b. tension in the guy wire [708],
c. horizontal and vertical forces exerted by the hinge on the beam [580,-6],
d. force exerted by the hinge on the beam [580, 0.593° below pos. x], and
e. force exerted on the wall by the hinge [same as d].
(For help, you may want to look at Example 9-6)
KEY:
The beam is in equilibrium. Use the conditions of equilibrium to calculate the
tension in the wire and the forces at the hinge. Calculate torques about the hinge,
and take counterclockwise torques to be positive.
Fg
m
g

155 N
 15.8163  15.8kg
9.80 m 2
s
FT
FH

   FT sin   l2  m1 g l1 2  m2 gl1  0 
1
2
FT 
m1 gl1  m2 gl1
l2 sin 

1
2
155 N 1.70 m    245 N 1.70 m 
1.35 m   sin 35.0o 
 708.0 N  7.08  102 N
F
F
x
 FH x  FT cos   0  FH x  FT cos    708 N  cos 35.0o  579.99 N  5.80  102 N
y
 FH y  FT sin   m1 g  m2 g  0 
FH y  m1 g  m2 g  FT sin   155 N  245 N   708 N  sin 35.0o  6.092 N  6 N  down 
FH on B  579.99 2   6.092 2  580.021  580 N
 6.092 


  tan 1 
  0.592693  0.593 below horizontal
579
.
99


FW on H  580 N at 0.593 below the horizontal
m1g
l1 2
l2
l1
m2g