A useful
methodology for analyzing statics problems is as follows:
1. Draw a free
body diagram – a free body must be a rigid object, i.e. one that cannot bend in
response to applied forces.
2. Draw all of
the forces acting on the free body. Is the number of unknown forces equal to the
total number of independent constraint equations shown in Table 1 (far right
column)? If not, statics can’t help you.
3. Decide on a
coordinate system. If the primary direction of forces is parallel and perpendicular
to an inclined plane, usually it’s most convenient to have the x and y coordinates
parallel and perpendicular to the plane, as in the cart and sliding block
examples below.
4. Decide on a
set of constraint equations. As mentioned above, this can be any combination of
force and moment of force balances that add up to the number of degrees of
freedom of the system (Table 1).
5. Decide on the
locations about which to perform moment of force constraint equations. Generally
you should make this where the lines of action of two or more forces intersect because
this will minimize the number of unknowns in your resulting equation.
6. Write down
the force and moment of force constraint equations. If you’ve made good choices
in steps 2 – 5, the resulting equations will be “easy” to solve.
7. Solve these
“easy” equations.
Example 1. Ropes
Two tugboats,
the Monitor and the Merrimac, are pulling a Peace Barge due west up Chesapeake Bay
toward Washington DC. The Monitor’s tow rope is at an angle of 53 degrees north
of due west with a tension of 4000 lbf. The Merrimac’s tow rope is at an angle
of 34 degrees south of due west but their scale attached to the rope is broken
so the tension is unknown to the crew.
 |
| Free Body Diagram of Monitor-Merrimac System |
a) What is the tension in the Merrimac’s
tow rope?
Define x as positive in the easterly direction, y as positive in the northerly direction. In order for the Barge to travel due west, the northerly pull by the Monitor and the Southerly pull by the Merrimac have to be equal, or in other words the resultant force in the y direction, Ry, must be zero. The northerly pull by the Monitor is 4000 sin(53°) = 3195 lbf. In order for this to equal the southerly pull of the Merrimac, we require FMerrimacsin(34°) = 3195 lbf, thus FMerrimac = 5713 lbf.
b) What is the
tension trying to break the Peace Barge (i.e. in the north-south direction)?
This is just the north/south force just computed, 3195 lbf
c) What is the
force pulling the Peace Barge up Chesapeake Bay?
The force exerted by the Monitor is 4000 cos(53°) = 2407 lbf. The force exerted by the Merrimac is 5713 cos(34°) = 4736 lbf. The resultant is Rx = 7143 lbf.
d) Express the
force on the Merrimac in polar coordinates (resultant force and direction, with
0° being due east, as is customary)
The magnitude of the force is 5713 lbf as just computed. The angle is -180° - 34° = -146°.
Example 2. Rollers
A car of weight
W is being held by a cable with tension T on a ramp of angle θ with respect to horizontal.
The wheels are free to rotate, so there is no force exerted by the wheels in
the direction parallel to the ramp surface. The center of gravity of the
vehicle is a distance “c” above the ramp, a distance “a” behind the front
wheels, and a distance “b” in front of the rear wheels. The cable is attached
to the car a distance “d” above the ramp surface and is parallel to the ramp.
 |
| Free body diagram for car-on-ramp with cable example |
(a) What is the
tension in the cable in terms of known quantities, i.e. the weight W,
dimensions a, b, c, and d, and ramp angle θ?
Define x as the direction parallel to the ramp surface and y perpendicular to the surface as shown. The forces in the x direction acting on the car are the cable tension T and component of the vehicle weight in the x direction = Wsinθ, thus ΣFx = 0 yields
Wsinθ - T = 0 ⇒ T = Wsinθ
(b) What are the
forces where the wheels contact the ramp (Fy,A and Fy,B)?
The forces in the y direction acting on the car are Fy,A,
Fy,B and component of the vehicle weight in the y
direction = Wcosθ. Taking moments of force about point A, that is ΣMA
= 0 (so that the moment of force equation does not contain Fy,A
which makes the algebra simpler), and defining moments of force as positive
clockwise yields
(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b)
– T(d) = 0
Since we already know from part (a) that T = Wsinθ, substitution
yields
(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b)
– (Wsinθ)(d) = 0
Since this equation contains only one unknown force, namely Fy,B,
it can be solved directly to obtain
Finally taking ΣFy = 0 yields
Fy,A + Fy,B
- Wcosθ = 0
Which we can substitute into the previous equation to find Fy,A:
Note the
function tests:
2) For θ = 0, T
= 0 (no tension required to keep the car from rolling on a level road)
3) As θ
increases, the tension T required to keep the car from rolling increases
4) For θ = 90°,
T = W (all of the vehicle weight is on the cable) but note that Fy,A
and Fy,B are non-zero (equal magnitudes, opposite signs)
unless c = d, that is, the line of action of the cable tension goes through the
car’s center of gravity.
5) For θ = 0, Fy,A
= (b/(a+b)) and Fy,B = (a/(a+b)) (more weight on the
wheels closer to the center of gravity.)
6) Because of
the – sign on the 2nd term in the numerator of Fy,A
(-(c-d)sin(θ)) and the + sign in the 2nd term in the numerator of Fy,B
(+(c-d)sin(θ)), as θ increases, there is a transfer of weight from the front
wheels to the rear wheels.
Note also that Fy,A
< 0 for b/(c-d) < tan(θ), at which point the front (upper) wheels lift
off the ground, and that Fy,A < 0 for a/(d-c) >
tan(θ), at which point the back (lower) wheels lift off the ground. In either
case, the analysis is invalid. (Be aware that c could be larger or smaller than
d, so c-d could be a positive or negative quantity.)
Example 3. Friction
A 100 lbf acts
on a 300 lbf block placed on an inclined plane with a 3:4 slope. The
coefficients of friction between the block and the plane are μs =
0.25 and μd = 0.20.
a) Determine
whether the block is in equilibrium
b) If the block
is not in equilibrium (i.e. it’s sliding), find the net force on the block
c) If the block
is not in equilibrium, find the acceleration of the block
 |
| Free body diagram for sliding-block example |
(a) To maintain
equilibrium, we require that ΣFx = 0 and ΣFy = 0.
Choosing the x direction parallel to the surface and y perpendicular to it,
ΣFy = N – (4/5)(300
lbf) = 0 ⇒ N = 240 lbf
so the maximum
possible friction force is
Ffriction, max
= μsN = 0.25 * 240 lbf = 60 lbf.
The force needed
to prevent the block from sliding is
ΣFx = 100 lbf –
(3/5)(300 lbf) + Fneeded = 0
Fneeded = -100 lbf + (3/5)(300 lbf) = 80 lbf
Which is more than the maximum available friction force, so the block will slide down the plane.
(b) The sliding
friction is given by
Ffriction, max
= μdN = 0.20 * 240 lbf = 48 lbf
so the net force
acting on the block in the x direction (not zero since the block is not at
equilibrium) is
ΣFx = 100 lbf –
(3/5)(300 lbf) + 48 lbf = -32 lbf
(c) F = ma ⇒ -32 lbf = 300 lbm * acceleration
acceleration = (-32 lbf/300 lbm)
???
what does this
mean? lbf/lbm has units of force/mass, so it is an acceleration. But how to
convert to something useful like ft/s2? Multiply by 1 in the funny form of gc =
1 = 32.174 lbm ft / lbf s2, of course!
acceleration = (-32 lbf/300 lbm)
(32.174 lbm ft / lbf s2) = -3.43 ft/s2
or, since gearth
= 32.174 ft/s2,
acceleration = (-3.43 ft/s2)/(32.174
ft/s2gearth) = -0.107 gearth.
The negative sign
indicates the acceleration is in the –x direction, i.e. down the slope of
course.
A good function
test is that the acceleration has to be less than 1 gearth, which is what you
would get if you dropped the block vertically in a frictionless environment.
Obviously a block sliding down a slope (not vertical) with friction and with an
external force acting up the slope must have a smaller acceleration.
Example 4. Rollers and Friction
A car of weight
W is equipped with rubber tires with coefficient of static friction μs.
Unlike the earlier example, there is no cable but the wheels are locked and
thus the tires exert a friction force parallel to and in the plane of the ramp
surface. As with the previous example, the car is on a ramp of angle θ with
respect to horizontal. The center of gravity of the vehicle is a distance “c”
above the ramp, a distance “a” behind the front wheels, and a distance “b” in
front of the rear wheels.
 | ||
Free body diagram for car-on-ramp with
friction example
|
(a) What is the minimum μs required to keep
the car from sliding down the ramp?
The unknowns are
the resulting forces at the wheels (Fy,A and Fy,B)
and the coefficient of friction μs. Taking ΣFx = 0, ΣFy
= 0 and ΣMA = 0 yields, respectively,
Note the function tests
1) For θ = 0, μs
= tan(θ) = 0 (no friction required to keep the car from sliding on a level
road)
2) As θ increases, the
friction coefficient μs required to keep the car from sliding increases
3) For θ = 0, Fy,A
= (b/(a+b)) and Fy,B = (a/(a+b)) (more weight on the
wheels closer to the center of gravity
4) Because of the – sign
on the 2nd term in the numerator of Fy,A (-c sin(θ)) and
the + sign in the 2nd term in the numerator of Fy,B (+c
sin(θ)), as θ increases, there is a transfer of weight from the front wheels to
the rear wheels.
Note also that we could have also tried ΣFy
= 0, ΣMA and ΣMB = 0:
Fy,A + Fy,B -
Wcosθ = 0
(Wsinθ)(c) + (Wcosθ)(a) – Fy,B (a+b) =
0
(Wsinθ)(c) - (Wcosθ)(b) + Fy,A (a+b) =
0
In which case, the second
equation could have been subtracted from the third to obtain:
Fy,A + Fy,B
- Wcosθ = 0
which is the same as the first
equation. So the three equations are not independent of each other, and we
can’t solve the system. What’s wrong? The coefficient of friction µs
doesn’t appear in the set of equations ΣFx = 0, ΣMA and
ΣMB = 0. We need to have each of the three unknowns Fy,A
, Fy,B and µs in at least one of the three
equations. The set ΣFx = 0, ΣMA and ΣMB = 0 doesn’t
satisfy that criterion.
(b) At what angle will the car
tip over backwards, assuming that it doesn’t start sliding down the ramp at a
smaller angle due to low µs ?
This will occur when Fy,A
= 0, i.e. when sin(θ)/cos(θ) = tan(θ) = b/c. This is reasonable because the
tip-over angle should increase when c is made larger (center of gravity closer
to the ground) or b made smaller (center of gravity shifted forward). Notice
also that it doesn’t depend on µs , that is, as long as it doesn’t
slide due to low µs , the tip-over angle only depends on the force
balance.
For what it’s worth, also note
that the tip-over angle equals the sliding angle when tan(θ) = µs =
b/c. Since generally µs < 1, except for a very top-heavy (large
c) or rear-weight-shifted (small b) vehicles, the vehicle will slide down the
ramp before it flips over backwards.
Example 5. Pinned Joint
 |
| Free body diagram for pinned joint example |
A straight bar of negligible mass 12 inches long is pinned at its lower end (call it point A) and has a roller attached to its upper end (call it point B) as shown in the figure. The bar is at a 30˚ angle from horizontal. A weight of 100 lbf is hung 4 inches from the lower end (call it point C).
a) What are the forces in the x and y directions on the pinned end? What is the force in the x direction on the roller end?
The pinned end can sustain forces in both the x and y directions, but no moment of force. The roller end can sustain a force only in the x direction, and again no moment of force. Summing the forces in the y direction
Fy,A + Fy,B + Fy,C
= 0 ⇒ Fy,A + 0 - 100 lbf = 0 ⇒ Fy,A = +100 lbf.
In other words, in the y direction the vertical force at point A must be +100 lbf since that is the only force available to counteract the 100 lbf weight. Next, taking moments of force about point A (since the lines of action of two of the unknown forces intersect at point A),
ΣMA = 0 ⇒ +(4
in)(cos(30˚))(100 lbf) + (6 in)Fx,B = 0 ⇒ Fx,B = -57.7 lbf.
Finally, for force balance in the x direction,
Fx,A + Fx,B + Fx,C
= 0 ⇒ Fx,A = -Fx,B - Fx,C
= -(-57.7 lbf) – 0 = +57.7 lbf
b) Would the forces change if the roller and pinned ends were reversed?
In this case summing the forces in the x direction:
Fx,A + Fx,B + Fx,C
= 0 ⇒ 0 + Fx,B + 0 = 0 ⇒ Fx,B = 0.
For force balance in the y direction,
Fy,A + Fy,B + Fy,C
= 0 ⇒ Fy,A + Fy,B = 100 lbf
Taking moments of force about point
C just for variety (not the easiest way, since neither Fy,A
nor Fy,B are known, we just know that Fy,A
+ Fy,B = 100 lbf),
ΣMC = 0 ⇒ (4 in)(cos(30˚))Fy,A - (8 in)(cos(30˚))Fy,B
+ (8 in)(sin(30˚))Fx,B = 0
⇒ (4 in)(cos(30˚))(100 lbf – Fy,B
) - (8 in)(cos(30˚))Fy,B + 0 = 0
⇒ Fy,B = +33.3 lbf ⇒ Fy,A = +66.7 lbf
which is quite different from case (a).
c) What would happen if the lower end were fixed rather than pinned (upper end having the roller again)?
In this case there are 4 unknown
quantities (Fx,A , Fy,A , MA
and Fx,B ) but only 3 equations (ΣFx = 0, ΣFy
= 0, ΣM = 0) so the system is statically indeterminate. If one takes away the roller
end entirely, then obviously Fy,A = 100 lbf, Fx,A
= 0 and MA = +(100 lbf)(4 in)(cos(30˚)) = 346.4 in lbf.
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