Equilibrium: Force and Mg

The First and Second Conditions for Equilibrium
The first condition for equilibrium: The second condition for equilibrium: • • ΣF = 0 ΣΓ = 0

•

In when both of these conditions are satisfied in static systems all forces and torques sum to zero. In problems where the first and second conditions of equilibrium are satisfied, the best strategy is to create FBD’s for both the first and second conditions, derive equations based on these FBD’s and then see what useful information may be gleaned from these equations. When applying the second condition we are free to choose any axis about which to compute torques. It is best to choose an axis that eliminates one or more forces that have lines of force that pass through it.

Example 1 Consider a playground seesaw. The mass of the plank is 2.0 kg, the masses of two children on it are 25 kg and 30 kg with the 30 kg child sitting 2.5 meters from the center of the plank (the fulcrum) as shown below. Where must the second child sit in order for this system to be in equilibrium?

Noting that a normal force directed upwards acts at the point of the fulcrum, the FBD’s for the first condition yield:

∑F

y

= N − mc1 g − m p g − mc 2 g = 0 →

∑F

y

= N − 294 N − 19.6 N − 245 N = 0

Note that while this is all true it is not, by itself, particularly useful. To apply the second condition we must first choose an axis about which to compute torques. The axis that makes the most physical sense would be one directly through the board over the fulcrum, but we could choose any axis that made computations easier. In this case choosing the axis associated with the fulcrum eliminates the forces created by the mass of the board itself since these act on the center of mass of the board which is located directly over the fulcrum.

The FBD’s for the second condition yield:

∑ Γ = (294 N )(2.5meters) − (245 N )( xmeters) = 0
Solving this equation for x yields a distance of 3 meters.

Example 2 Consider the following cantilevered beam:

The beam has a mass of m = 25 kg and is 2.2 meters long. The suspended block has a mass M = 280 kg and the supporting cable makes an angle of 300 with the beam. Determine the force that the wall exerts on the beam at the hinge and determine the tension in the supporting cable.

• •

Notice that the normal force is the x component of the force exerted by the wall on the beam through the hinge (Fx). Because the beam is also held up by the hinge (Fy) the total force the wall exerts on the beam is the aggregate of these two components. So we must determine, from the available information, Fx, Fy, Tx, Ty and finally T and F.

Application of the first condition with our sign convention yields:

∑F ∑F

x

= Fx − Tx = 0 ∴ Fx = Tx = Fy + T y − mg − Mg = 0

y

Application of the second condition with respect to the hinge yields:

Fy – line of action passes through the hinge → no torque Fx – line of action passes through the hinge → no torque Tx – line of action passes through the hinge → no torque mg – exerts a torque Mg – exerts a torque Ty – exerts a torque
With our sign convention:

∑ Γ = −(mg )(1.1m) − (Mg )(2.2m) + (T

y

)(2.2m) = 0 )(2.2m) = 0

∑ Γ = −(245 N )(1.1m) − (2744 N )(2.2m) + (T

y

∑ Γ = −270 N ⋅ m − 6037 N ⋅ m + (T v y

)(2.2m) = 0 → T y = 2867 N

Since T y = T sin θ : T = 5734 N , and with a little more work Tx = 4966 N . With the magnitude of T and all of its components known, it is a simple matter to substitute into the equation in y from the first condition and solve for Fy:

∑F

y

= Fy + 2867 N − 245 N − 2744 N = 0

Fy = 122 N
Noting that Fx = Tx (why?), Fx = 4966N → F =

(4966 N ) 2 + (122 N ) 2 = 4967 N

tan o = /

y 122 N = ∴ o = 1.4° / x 4966 N

v F = 4967 N @ 1.4°

Example 3. A 5 meter long ladder leans against a frictionless wall. The point of contact between the ladder and the wall is 4 meters above the ground. The ladder is uniform with a mass of 12 kg. Determine the forces exerted by the ground and wall on the ladder.

A little yields θ = 53° and x = 3m. FBD

Notice that since the wall is frictionless the force that it exerts on the latter is normal to the surface of the wall. It is necessary to find the component perpendicular to the ladder only for the purpose of computing a torque. The force that the ground exerts on the ladder, however, does have two components. Can you explain why? Applying the first condition yields:

∑F

y

= Fgy − mg = 0

Fgy = mg ∴ Fgy = 118 N

∑F

x

= Fgx − Fwall = 0

Fgx = Fwall
Applying the second condition with respect to the point of contact between the ground and latter (this eliminates Fg and it’s components from torque computations):

∑ Γ = (F

wall

)(sin θ )(5m) − (mg )(cosθ )(2.5m) = 0 )(4m) − (71N )(2.5m)

∑ Γ = (F

wall

Fwall = 44 N
Recall that Fgx = Fwall ∴ Fgx = 44 N and:

Fg = (44 N ) 2 + (118 N ) 2 = 126 N

tan o = /

y 118 N = ∴θ = 70° x 44 N

v Fg = 126 N @ 70°

Now lets use the same ladder but with a 60kg painter 3.5 meters up the ladder and determine what the coefficient of friction, µ, is between the ladder and the floor.

Fwsinθ Fw

Fwsinθ is the component of Fw perpendicular to the ladder

Mg FGy mg θ FGx Pivot mgcosθ

Mgcosθ

Mgcosθ is the component of Mg perpendicular to the ladder

mgcosθ is the component of mg perpendicular to the ladder

+

∑F
∑F

x

= FGx − Fw = 0 ∴ FGx = Fw = FGy − mg − Mg = 0
Note: FGx = µFGy

y

FGy = (m + M ) g = (12kg + 60kg ) 9.8m ⋅ s −2 FGy = 706 N

(

)

(How does this compare to the earlier result?)

Recall: θ = 53°

∑ Γ = (F
Fw =

w

)(sin θ )(5m) − ( Mg )(cos θ )(3.5m) − (mg )(cos θ )(2.5m) = 0

1239 N ⋅ m + 178 N ⋅ m = 354 N (How does this compare to the earlier result?) 4m Fw = .5 FGy

Fw = µFGy ∴ µ =

Example 4. A system consists of a ladder length ℓ and mass m on a rough floor leaning against a smooth vertical wall. A man of mass M is standing 1/3 of the way up the ladder from the floor. His dog, Bantor, stands on the ladder at its foot. The angle between the ladder and the floor is θ. Draw a free body diagram for this system and express the first and second conditions of equilibrium in terms of ℓ and θ about an axis at the base of the ladder.

N2

N2sinθ θ

θ

ℓ

N1

Ffloor

mg

θ mg θ O θ Mg Mgcosθ mgcosθ

o Mg / θ Bantor f mbantor

1st Condition:

∑F ∑F

x y

= f − N2 = 0 = N 1 − Mg − mg − m bantor g = 0

Bantor’s mass exerts no torque since he stands at the origin 2nd Condition:

∑Γ = N

2

⎛l⎞ ⎛l⎞ sin θ (l ) − mg cos θ ⎜ ⎟ − Mg cos θ ⎜ ⎟ = 0 ⎝ 3⎠ ⎝2⎠

Example 5. The maintenance staff in a hockey arena wishes to hang a scoreboard on the wall. They use a ladder with a uniform mass of 25kg to do the job. The ladder is 10 meters in length and an 80kg worker needs to be 7/10 of the way up the ladder to hang the scoreboard. The base of the ladder rests on a very smooth (icy) surface. If the coefficient of friction between the top of the ladder and the wall is 0.7, and the angle the ladder makes with the ice is 70°, what force must an assistant supply (who is presumably standing on a non-slippery surface) to the base of the ladder so that it does not slip?

fcosθ and Nsinθ f
N θ

mgcosθ
Fn Fapp 70°

Mgcosθ Mg

mg

1st condition:

∑F ∑F

x

= Fapp − N = 0 ∴ Fapp = N = Fn − mg − Mg + f = 0 Q Fn + µN − mg − Mg = 0

y

2nd condition:

∑ Γ = (−mg cosθ )(5m) − (Mg cosθ )(7m) + ( N sin θ )(10m) + ( f cosθ )(10m) = 0
)(cos 70°)(5m) + (80kg )(9.8m ⋅ s −2 )(cos 70°)(7m) = (( N )(sin 70°)(10m) ) + (µN (cos 70°)(10m) )
Note: N = Fapp (from the 1st condition)

((25kg )(9.8m ⋅ s

−2

) (

)

419 N ⋅ m + 1877 N ⋅ m = Fapp (9.39 + 2.4)m

Fapp = 195 N

Example 6. A uniform ladder is leaning against a smooth, vertical wall. The ladder makes an angle of θ with the ground. If the coefficient of friction between the ladder and the ground is 0.70, find the minimum value of θ such that the ladder does not slip.

N

Nsinθ

Fn f
1st condition:

mgcosθ mg

∑F ∑F

x

= f −N =0 Q

f = N = µFn = µmg

y

= Fn − mg = 0 ∴ Fn = mg

2nd condition:

∑ Γ = −mg cosθ ⎜ 2 ⎟ + N sin θ (l ) = 0 Q mg cosθ ⎜ 2 ⎟ = µmg sin θ (l ) ⎝ ⎠ ⎝ ⎠
⎛1⎞ cos θ ⎜ ⎟ = µ sin θ ⎝2⎠ 1 = tan θ 2µ

⎛l⎞

⎛l⎞

θ = 35.5°

Example 7. A ladder (m = 1.20kg) leans against a frictionless wall 4.0 meters above the ground. The base of the ladder is 3.0 meters from the wall. A 60kg painter stands 70% of the way up the ladder. Assuming that the ladder is stable what is the minimum coefficient of static friction between the ladder and the ground?

Fw 3.5m 2.5m Fy θ mg Fx

5m Mg

Use the Pythagorean Theorem to solve for the length of the ladder:

a 2 + b 2 = c 2 ∴ c = 5m
Noting that this is a 3 − 4 − 5 triangle θ ≈ 53° 1st Condition:

∑F

x

= Fx − Fw = 0 ∴ Fx = Fw = Fy − mg − Mg = 0 ∴ Fy = (m + M ) g = (1.20kg + 60kg ) 9.8m ⋅ s −2 = 600 N

∑F

y

(

)

Note: Fx = f s = µFy ∴ µ (600 N ) = Fw 2nd Condition:

∑ Γ = −mg cosθ (2.5m) − Mg cosθ (3.5m) + F

w

sin θ (5m) = 0

mg cos θ (2.5m) + Mg cos θ (3.5m) = µ (600 N )sin θ (5m)

(1.20kg )(9.8m ⋅ s −2 ) cos(53°)(2.5m) + (60kg )(9.8m ⋅ s −2 ) cos(53°)(3.5m) µ= = 0.52 (600 N )sin (53°)(5m)