Example 1
A robot has a mass of 60 kg. How much does that robot weigh
sitting on the earth at sea level?
Given: mRobot = 60 kg
Find: WRobot
Relationships: W = mg , g = 9.81 m/s2
Solution:
WRobot = mRobot g = 60 ( 9.81) = 589 N
⇒ WRobot = 589 N
Example 2
A robot designed for lunar exploration has a mass of 60 kg.
Given the moon has a gravitational acceleration constant of
gMoon = 1.62 m/s2 , how much will the robot weigh sitting on the
surface of the moon?
Given: mRobot = 60 kg
Find: WRobot-Moon
Relationships: W = mg , gMoon = 1.62 m/s2
Solution:
WRobot-Moon = mRobot gMoon = 60 (1.62) = 97.2 N
⇒ WRobot-Moon = 97.2 N
Example 3
Two identical metal blocks, A and B, are hanging from a rod.
Cable EF holds block B to block A, and cable CD holds block A
to the rod. Each block weighs 10 pounds, and the system is in
static equilibrium. Find the tension in the two cables.
C
D
A
Given: WA = WB = 10 lb ; the system is in static equilibrium
⇒
→
∑F = 0 .
E
F
Find: Tensions TCD and TEF .
Relationships:
B
→
∑F = 0
Solution:
Since there are fewer forces acting on the bottom block, block B,
begin the analysis there. As with so many other things, its easier to
start with something simple and work towards the harder than the
other way around.
Isolate block B. Note that there are two forces acting on that block,
the block's weight, and the tension in cable EF.
Since the system is in static equilibrium, block B is in static
equilibrium. This means the sum of all the forces acting on block B is
equal to zero.
→
∑ F Acting on B = 0
⇒
→
∑F
Acting on B
= ∑ Fy = TEF − WB = 0
Note:
• All the forces acting on block B act in the y direction. We will still
be using vector math, but with only one dimension, it will appear
to be simple algebra.
• According to the frame of reference I choose, tension TEF acts in
the positive y direction; the weight of block B acts in the negative
y direction. Hence, the equation TEF − WB .
y
x
TEF
B
WB
Solving the equation
∑F
y
= TEF − WB = 0
⇒ TEF = WB = 10 lb
⇒ TEF = 10 lb
Now, isolate block A. Note that there are three forces acting on that
block, the block's weight, the tension in cable CD, and the tension in
cable EF.
The weight of block B does not act on block A, only the tension of
cable EF.
Since the system is in static equilibrium, block A is in static
equilibrium. This means the sum of all the forces acting on block A is
equal to zero.
→
∑F
⇒ ∑F
Acting on A
=0
→
Acting on A
= ∑ Fy = TCD − W A − TEF = 0
Solving the equation
y
TCD
= TCD − W A − TEF = 0
⇒ TCD = W A + TEF = 10 + 10 = 20 lb
⇒ TCD = 20 lb
x
A
WA
TEF
Again, all the forces acting on block A act in the y direction.
∑F
y
Example 4
y
A support member of a robotic arm is
loaded as shown. The member is made of a
light weight composite such that it's weight
can be neglected ( WMember = 0 ). We know
from other analysis that the force
Ax = 77 N , and force B = 200 N .
Determine the magnitude of forces
Ay and C . The system is in static
x
Ax
Ay
30°
B
2m
C
4m
equilibrium.
Given: Ax = 77 N, B = 200 N, WMember = 0 ; the system is in static equilibrium ⇒
→
∑F = 0 .
Find: Magnitudes Ay and C .
Relationships:
→
∑ F = 0 , sinθ =
opposite
adjacent
, cosθ =
hypotenuse
hypotenuse
Solution:
y
First, note that all of the forces shown lie parallel to the given frame of
→
x
reference with the exception of C . The first thing we should do in this
→
problem is determine the components of vector C that lie in the x and y
directions. This will simplify our calculations later.
30°
C
opposite
sinθ =
⇒ sin ( 30o ) = x ⇒ Cx = C sin ( 30o )
hypotenuse
C
C
adjacent
cosθ =
⇒ cos ( 30o ) = y ⇒ Cy = C cos ( 30o )
hypotenuse
C
By completing this calculation, all the force vectors we are working with lie in
either the x or y direction. This means we can work the problem in those two
directions separately.
y
x
Cx
30°
→
Additionally, decomposing vector C into its components did not add an
additional unknown. There is still only one unknown here, namely the
magnitude of the vector, C . The sine and cosine functions are just scalar
multiples of the magnitude of vector C .
C
Cx
C
y
→
After decomposing vector C , the system
simplifies to the following
x
Ax
Cx
Ay
Cy
B
2m
4m
Given the system is in static equilibrium, we merely need to sum forces in the x direction and set them
equal to zero, then sum forces in the y direction and set them equal to zero.
Incidentally, there is no magic in the order in which one sums equations. One could sum y first, then
x.
However, do not mix y components with x components!
To continue …
∑F = 0
⇒ ∑F = A
x
x
x
− Cx = 0
⇒ Ax = Cx = 77 N
⇒ C = 154 N
⇒ Cx = 77 = C sin ( 30o )
⇒ C=
77
= 154 N
sin ( 30o )
Then …
∑F
y
⇒
=0
∑F
y
= Ay − B + Cy = 0
(
)
⇒ Ay − 200 + C cos ( 30o ) = 0
(
)
⇒ Ay = 200 − 154cos ( 30o ) = 66.6 N
⇒ Ay = 66.6 N
Example 5
y
Block A has a mass of 5 kg. The spring has an unstretched length of
0.25 m ( LO = 0.25 m ). The system is in static equilibrium in the position
shown to the right. Determine the spring constant, k, of the spring.
x
350 mm
Given: mA = 5 kg, LO = 0.25 m ;
the system is in static equilibrium ⇒
→
∑F = 0 .
A
Find: k .
Relationships:
→
∑F = 0 , F
S
= k L − LO , W = mg
Solution:
Since there are only two forces acting on block A, begin the analysis there.
FS
First, determine the weight of block A
W A = m A g = ( 5 )( 9.81) = 49.05 N
x
A
Since the system is in static equilibrium, the sum of all the forces
acting on block A is equal to zero.
→
∑F
⇒
Acting on A
=0
→
∑ F Acting on A = ∑ Fy = FS − WA = 0
⇒ FS = W A = 49.05 N
To determine the spring constant, we will use the spring equation.
However, watch that you are consistent with units, in this case units of length.
L = 350 mm = 0.35 m
Then
FS = k L − LO ⇒ k =
y
FS
49.05 N
=
= 490.5 N m ⇒ k = 490.5 N m
L − LO
0.35 − 0.25 m
WA
Example 6
→
A robot will apply a force, F , to hold the 6 kg block in place. The
→
coefficient of static friction is μS = 0.4 . Determine the largest force, F ,
the robot can apply for which the box will not slip.
→
F
30°
Given: mBlock = 6 kg, μS = 0.4 ;
the system is in static equilibrium ⇒
→
∑F = 0 .
→
Find: Largest force F that can be applied without causing box to slip.
Relationships:
→
∑ F = 0 , f = μSN , W = mg
Solution:
y
This is a friction problem. Since the block will be about to slip or move, we
will determine the normal force, use it to determine the full force of static
x
WBlock
→
friction, and finally determine force F . That is the strategy.
Note the following with regard to the diagram on the right:
•
The force of friction points down the ramp. Correctly finding the direction
of friction is key in correctly solving these type of problems. In this case
→
(as always), friction is NOT opposing F , but opposing the motion of the
→
•
block. If we are looking for the largest force F , then that force will tend
to cause the block to move up the ramp. Since friction opposes motion, it
will oppose the block moving up the ramp. Hence, friction points down
the ramp.
The block remains in its original orientation, but the frame of reference
has been skewed 30°.
o Never change the orientation of the object you are analyzing. We
are accustomed to drawing the force of weight straight down, for
that is where the center of the earth is. If you skew the orientation
of the object, the weight vector will not point straight down.
However, most people will forget and incorrectly draw that vector.
o All of the forces in the diagram lie parallel or perpendicular to the
ramp with the exception of the weight vector. It is quicker to reorient the frame of reference and resolve the weight vector into
components, than to use a horizontal/vertical frame and resolve
three vectors into their components.
F
f
N
Decompose the weight vector into its components with regard to the given
frame of reference.
sin ( 30o ) =
Wx
⇒ W x = WBlock sin ( 30o )
WBlock
cos ( 30o ) =
Wy
y
⇒ Wy = WBlock cos ( 30o )
WBlock
x
WBlock
The force of weight
WBlock = mBlock g = 6 ( 9.81) = 58.86 N
F
N
f
⇒ W x = WBlock sin ( 30o ) = 58.86sin ( 30o ) = 29.43 N
Wy = WBlock cos ( 30o ) = 58.86cos ( 30o ) = 50.97 N
y
Wx
30° WBlock
The force of friction (recall, motion is impending)
Wy
x
f = μSN = 0.4N
The system is in equilibrium, so
→
y
∑ F = 0 . Note that only two forces lie in the
x
y direction, so start there. The solution should be reached quicker.
Wy
∑ Fy = 0
⇒
∑F
y
Wx
= N − Wy = 0
⇒ N = Wy = 50.97 N
⇒ N = 50.97 N
∑F = 0
⇒ ∑F = F −W
x
x
x
−f = 0
⇒ F = W x + 0.4N
⇒ F = 29.43 + 0.4 ( 50.97 ) = 49.82 N
⇒ F = 49.8 N
F
f
N
Example 7 – The remaining examples are for you to complete.
A robot must be able to push a block with a mass of 80 kg.
What is the weight of the block at sea level?
Given: A box has a mass of 80 kg.
Find: The weight of the box.
Relationships: W = mg , g = 9.81 m/s2
Solution:
Answer: W = 785 N
Example 8
A robot must be able to push a block with a mass of 80 kg on
the moon. What is the weight of the block on the surface of
the moon? The gravitational constant for the moon is
gMoon = 1.62 m/s2 .
Given: A box has a mass of 80 kg.
Find: The weight of the box on the moon.
Relationships: W = mg , gMoon = 1.62 m/s2
Solution:
Answer: W = 130 N
Example 9
A 60 lb cylinder rests in the crevasse of two boards, inclined
as shown. What are the normal forces exerted by the right and
left board on the cylinder?
Given: A cylinder weighing 60 lb.
30o
45o
Find: The normals NLeft and NRight .
Relationships: sinθ =
opposite
adjacent
, cosθ =
hypotenuse
hypotenuse
Answer: NRight = 31.1 lb
NLeft = 43.9 lb
Solution: For this example, we'll give you a starting hint, the FDB of the cylinder.
y
x
W
NLeft
60°
45°
NRight
Example 10
y
A 2 kg block is resting on a smooth surface inclined 30° as shown. It is
held off another surface by a spring with a spring constant of
k = 50 N m . How far is the spring compressed?
Given: mBlock = 2 kg , k = 50 N m .
Find: The amount of compression δ .
Relationships: W = mg , FSpring = k δ , g = 9.81 m s2
Solution:
x
30°
Answer: δ = 0.196 m