


Conditions for Static Equilibrium
Worksheet 1  Questions and Answers on Static Equilibrium

1. An object of mass, M, is suspended between a block of mass 10kg and a wall by a string which makes and angle of 30°with the vertical. The coefficient of friction between the table top and the block is 0.5. What is the maximum weight that can be suspended?

Fig 1. 
Answer 


At X the forces are:
T_{1} at 60° to the horizontal
T_{2L}  horizoontal
W  vertical
W = Mg


Resolving T_{1} into vertical and horizontal components
Vertical component = T_{1} Sin 60°
Horizontal component = T_{2L}Cos 60° 

Forces on 10kg block at Y
R = Normal reaction
Weight of block = mg
Weight of block = 100N
For equilibrium R = 100N
T_{2R} = Tension on the string attached to the block
Tension on the horizontal string is constant so the block experiences a force of T_{2R} = T_{2L}. 
Fig 1b. 



Since X and Y are in equilibrium:
At Y
∑F = 0
∑Horizontal forces = 0
∑Vertical forces = 0
Normal reaction of table top = Weight of M
Normal reaction of table top (R) = 100N
Maximum friction (T_{2R}) = μR
(T_{2R}) = (0.5) (100)
(T_{2R}) = (50N) Eq1
Tension on the horizontal string is constant so the block experiences a force of T_{2}.

Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— The linear momentum ()of its center of mass is zero.

Vertical forces: (using up as positive)
At X
∑Vertical forces = 0
T_{1} Sin 60° + (W) = 0
T_{1} Sin 60° = W
T_{1} Sin 60° = mg
m = (T_{1} Sin 60° / 10)  Eq2

Recall that we can treat with horizontal and vertical forces independently as a force has no component at right angles to it so vertical forces can have no horizontal forces and viceversa.

Horizontal forces: (using right as positive)
At X
∑Horizontal forces = 0
T_{2L} + T_{1} Cos 60°= 0
(50N) + T_{1} Cos 60°= 0 (T_{2} points to the left)
(50N) = T_{1} Cos 60°
(50) = 0.5 T_{1}
T_{1} = 100N Eq3
Substituting for T_{1} in Eq2 from Eq3
m =
[(100N Sin 60°/ 10)]_{}
m = 8.6 kg

Alternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.
Tan 60° = ^{W}/_{T2L}
W= ^{}[(T_{2L}) X (Tan 60°)]
W = _{}86N
W = Mg
M  8.6 kg 








2. A leaf of mass 1 gram hangs on a rafter by a spider's thread as in fig 2b. The wind blows horizontally and the thread makes an angle of 20° with the vertical. Calculate (a) the tension on the thread and (b) the force of the wind on the leaf.

Fig 2a. 
Answer 


Forces on the Leaf [Fig 2 (i)]
Let T = Tension on the web
F = Force of the wind
W = Weight of the leaf
Weight of leaf (W) = mg
W = (1 X 10^{3} kg) (10ms^{2})
W = (1 X 10^{2} N)
Resolving T into vertical and horizontal components [Fig 3b(ii)]
Vertical Component = T Cos 20°
Horizontal component = T Sin 20°

Fig 2b. 

(a) ∑F = 0
∑Vertical forces = 0
T Cos 20° + (W) = 0
T Cos 20° =  (W)
T Cos 20° = (1 X 10^{2} N) Eq1
T = (1 X 10^{2} N) / Cos 20°
T = (1.1 X 10^{2} N)
∑Horizontal forces = 0
(T Sin 20°) + F = 0 Eq2
F = T Sin 20°
F = (3.6 X 10^{3} N)

Double check Answer.
T Sin 20° = (1 X 10^{2} kg) Eq1
T Cos 20° =  F Eq2 rearranged.
Dividing Eq1 by Eq 2
Tan 20° = (1 X 10^{2} kg) / ( F)
( F) = (1 X 10^{2} kg) / Tan 20°
F = (2.7 X 10^{2} N)
Alternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.
Cos 20° = ^{W}/_{T}
T = ^{W}/_{Cos 20°}
T = ^{(0.01N)}/_{Cos 20°}
T = (1.1 X 10^{2} N)
Tan 20° = ^{F}/_{W}
F = ^{Tan 20°}/_{W}
F = ^{(Tan 20°)}/_{(0.01N)}
F = (3.6 X 10^{3} N)
Double check:
(1.1 X 10^{2} N)^{2} = (3.6 X 10^{3} N)^{2} +
(1.0 X 10^{2} N)^{2}









3. A flower pot of mass 8kg is hung from two wires as show in in fig 3a. Find the tension in each wire.

Fig 3a. 
Answer 


Forces on the junction
T_{1} = Tension on the right wire
T_{2} = Tension on the left wire
W = Weight of the flower pot
Weight of flower pot (W) = mg
W = (8^{} kg) (10ms^{2})
W = (80^{} N)


Resolving T_{1} into vertical and horizontal components
Vertical Component = T_{1} Sin 40°
Horizontal component = T_{1} Cos 40°
Resolving T_{2} into vertical and horizontal components
Vertical Component = T2_{1} Sin 60°
Horizontal component = T_{2} Cos 60°

Fig 3b. 

(a) ∑F = 0
∑Horizontal forces = 0
T_{1} Cos 40° + [(T_{2} Cos 60°)] = 0
T_{1} = [(T_{2} Cos 60°) / (Cos 40°)]
T_{1} = 0.65 T_{2} Eq2
∑Vertical forces = 0
T_{1} Sin 40° + T_{2} Sin 60° + (W) = 0
T_{1} Sin 40° + T_{2} Sin 60° = (80N)
T_{1} Sin 40° + T_{2} Sin 60° = (80N)
T_{1} Sin 40° = (80N)  T_{2} Sin 60°
T_{1} = {[(80N) / (Sin 40°)  (T_{2} Sin 60°)] / (Sin 40°)}
T_{1} = 124N  1.35 T_{2} Eq1
Substuting for T_{1} from Eq1 into Eq2
0.65 T_{2} = 124N  1.35 T_{2}
124N =
1.35 T_{2}
+ 0.65 T_{2}
124N =
2 T_{2}
T_{2} = 62 N
T_{1} = 41N

Alternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.
Using (^{W}/_{Sin 100° }) = (^{T1}/_{Sin 30°}) = (^{T2}/_{Sin 50°})
(^{80N}/_{Sin 100° }) = (^{T1}/_{Sin 30°}) = (^{T2}/_{Sin 50°})
T1 = ^{80 Sin 30°}/_{Sin 100°}
T1 = 41N
T2 = ^{80 Sin 50°}/_{Sin 100°}
T2 = 62N










4. Several men with a combined strength of 4000N pull horizontally on a girder of mass 5000kg suspended from a crane. Assuming that the rope stays horizontal during the operation, What is the maximum angle, θ, that the bean can be shifted?

Fig 4a. 
Answer 


Forces on the beam
T_{1} = Tension on the rope
T_{1} = 4000N
T_{2} = Tension on the cable from the crane
W = Weight of the beam
Weight of beam (W) = mg
W = (5000^{} kg) (10ms^{2})
W = (50000^{} N)


Resolving T_{2} into vertical and horizontal components
Vertical Component = T_{2} Cos θ
Horizontal component = T_{2} Sin θ

Fig 4b. 

(a) ∑F = 0
∑Horizontal forces = 0
T_{1} + [(T_{2} Sin θ)] = 0
4000N_{} = T_{2} Sin θ
Sin θ _{ }= 4000N / T_{2} Eq1
∑Vertical forces = 0
T_{2} Cos θ + (W) = 0
T_{2} Cos θ = W
T_{2} Cos θ = 50000N
Cos θ = 50000N / T_{2} Eq2
Dividing Eq1 and Eq2
Sin θ _{ }= 4000N / T_{2}
Cos θ = 50000N / T_{2}
Tan θ = (4000N / T_{2}) / (50000N / T_{2})
Tan θ = (4000N) / (50000N)
θ = 4.6°
Alternative Solution
Sin_{2} θ = (4000N)_{}_{2} / T_{22}) Eq4
Cos_{2} θ = (50000N)_{2}_{} / T_{22}) Eq4
Adding Eq1 and Eq2
Cos_{2} θ + Sin_{2} θ = (4000N)_{}_{2} / T_{22}) +(50000N)_{2}_{} / T_{22})
1 = [(4000N)_{}_{2}) + (50000N)_{2}]/ T_{22}
T_{22} = [(4000N)_{}_{2}) + (50000N)_{2}]
T_{2} = ± 50160N
From Eq2
Cos θ = 50000N / 50160N
θ = 4.6°
Doublecheck
From Eq1
Sin θ = 4000N / 50160N
θ = 4.6°

Alternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.
Tan θ = (^{T1}/_{W})
Tan θ = (^{4000N}/_{50000N})
θ = 4.6°









5. A 3m ladder having a uniform density and a mass 30kg rests against a frictionless vertical wall at an angle of 30°. The lower end rests on a flat surface where the coefficient of static friction is µ = 0.40. A painter with a mass M = 60kg is at the middle of the ladder. At what point in his journey up the ladder will it begin to slip?

Fig 5.1. 
Answer 


Forces on the ladder
F_{w} = Normal reaction of the wall
W_{L} = Weight of ladder acts at P (centre of the ladder)
W_{m} = Weight of man
F_{n} = Normal reaction of the ground
F_{f} = Frictional force on the ladder from the ground
Weight of ladder (W_{L}) = mg
W_{L} = (30^{} kg) (10ms^{2})
W_{L} = (300^{} N)
Weight of man (W_{m}) = mg
W_{m} = (60^{} kg) (10ms^{2})
W_{m} = (600^{} N)
z = Length of ladder X Sin 30°
z = 3m Sin 30°
y = (Length of ladder / 2) X Sin 30°
y = 1.5m Sin 30°
Στ = 0
Taking Moments about A
[
(F_{n} X 0m) + (F_{r} X 0m) +
[(W_{m} X xm)] +
[(W_{l} X ym)] +
(F_{w} X zm) = 0
[(600N_{} X xm)] +
[(300N_{} X1.5m Sin 30°m)] +
(F_{w} X 6m Cos 30°m) = 0
F_{w }={[(600N_{} X xm) +
(300N_{} X1.5m Sin 30°m)] / (3m Cos 30°m)}
F_{w }={[(600x) +
(225N)] / 2.6)}
F_{w }= (230x) +
(86N) Eq1
ΣF = 0
Vertical Forces
F_{n} + W_{m} + W_{L} = 0
F_{n} + [(600N)] _{} + [(300N)] _{} = 0
F_{n} = (600N) _{} + (300N)
F_{n} = 900N
Horizontal Forces
F_{f} + F_{w} = 0  Eq2
F_{f} = µF_{n}
F_{f} = 0.4 X 900N
_{}F_{f} = 360N
From Eq2
(360N)
_{} + (F_{w}) = 0
F_{w} = 360N
From Eq1
F_{w }={[(600N_{} X xm) +
(300N_{} X1.5m Sin 30°m)] / (3m Cos 30°m)}
360N_{} ={[(600N_{} X xm) +
(300N_{} X1.5m Sin 30°m)] / (3m Cos 30°m)}
(360N_{}) (3m Cos 30°m) = {[(600N_{} X xm) +
(300N_{} X1.5m Sin 30°m)]}
{[(360N_{}) _{} (3m Cos 30°m)  (300N_{} X1.5m Sin 30°m)] / (600N)} _{}= xm
x = 1.18m
Length of AP = x / Cos 60°
Length of AP = 2.36m
He can climb 80% of the ladder.








6. The system in Fig. 6.1 is in equilibrium with the string in the center exactly horizontal.
Find (a) tension T_{L}, (b) tension T_{h}, (c) tension T_{r} and (d) angle θ.

Fig 6.1. 
Answer 


Since X and Y are in equilibrium:
∑F = 0
At X
∑Horizontal forces = 0
[(T_{L} Sin 35°)] + T_{h} = 0 Eq1
∑Vertical forces = 0
(80.0N) + (T_{L} Cos
35°) = 0 Eq2
At Y
∑Horizontal forces = 0
T_{r} Sin θ + (T_{h}) = 0 Eq3
∑Vertical forces = 0
(100N) + T_{r} Cos
θ° = 0 Eq4
From Eq1
T_{h} = T_{L} Sin 35° Eq5
From Eq2
T_{L} = ^{(80.0N)}/_{(Cos 35°)}
T_{L} = 97.7N Eq6
Eq5 becomes T_{h} = (97.7N) Sin 35°
T_{h} = 56.0N
From Eq3
T_{r} Sin θ = T_{h}
T_{r} Sin θ = 56.0N Eq7
From Eq4
T_{r} Cos θ = 100N Eq8
Dividing Eq7 by Eq8
Tan
θ = ^{(56.0N)}/_{(100N)}
θ = 29.3 °
From Eq8
T_{r}= ^{(100.0N)}/_{(Cos 29.3 °)}
T_{r}= 114N








7. The crane in Fig. 7.1 is in equilibrium. A ball of mass 225 kg hangs from the end of the uniform strut whose mass is 50.0 kg.
Find (a) the tension T^{c} in the cable and the (b) horizontal and (c) force exerted on the strut by the hinge.

Fig 7.1. 
Answer 
Weight of Strut = 500N
Weight of ball = 2250kg
Length of Strut = L m
The Strut is in Equilibrium so
Στ
= 0
Taking moments about F:
(R_{} X 0) + (T_{c} X BC) + [(W_{b} X DF) + [(W_{s} X EF) = 0
(0) + (T_{c} X L Sin 15°) + [(2250_{} X L Cos 60°)] + [(500_{} X ½L Cos 60°)] = 0
(T_{c} X L Sin 15°) = (2250_{} X L Cos 60°) + (500_{} X ½L Cos 60°)
T_{c} = [(2250_{} X Cos 60°) + (50_{}0 X ½ Cos 60°)] / (Sin 15°)
T_{c} = 4830N
The Strut is in Equilibrium so
ΣF
= 0
Vertical forces
(F_{v}) + [(W_{b}) + [(W_{s})] + [(T_{c} Sin 45°)] = 0
(F_{}_{v}_{}) = [(W_{b}) + (W_{s})] + (T_{c}Sin 45°))
(F_{v}_{}_{}) = [(W_{b}) + (W_{s})] + (T_{c}Sin 45°))
(F_{v}_{}_{}) = [(2250N_{}) + (500N_{}) + (3415N_{})]
(F_{v}_{}_{}) = 6165N
Horizontal forces
(F_{h}_{} ) + [(T_{c} Sin 45 °)]] = 0
(F_{h}_{} ) = (T_{c} Sin 45 °)
(F_{h}_{}) = 3415N
F^{2} = F_{v}_{}^{2} + F_{h}_{}^{2}
F^{ } = 7048N
Tan Θ =
F_{v}_{}^{ } / F_{h}_{}^{ }
Θ = 61 °
F^{ } = 7048N at elevation 61 °.









8. The stepladder shown in Fig. 8.1, has sides AC and CE are each 3.0 m long with a hinge at C. The tie bar BD is 0.6m long, halfway up. An man of mass 80kg climbs 2.0m along the ladder. Assuming that the floor is frictionless and neglecting the weight of the ladder, find (a) the tension in the tie–rod and the forces exerted on the ladder by the floor at (b) A and (c) E.

Fig 8.1. 
Answer 









9. A 50 kg wooden crate rests on a wooden ramp with an adjustable angle of inclination and a coefficient of friction between the block and ramp of 0.30.
(a) Draw a free body diagram of the crate.
(b) If the angle of the ramp is set to 10°, determine…
(i) the component of the crate's weight that is perpendicular to the ramp
(ii)
the component of the crate's weight that is parallel to the ramp
(iii)
the normal force between the crate and the ramp
(iv) the static friction force between the crate and the ramp
(c) At what angle will the crate just begin to slip?

Answer (a) 


W = Weight of crate
W = mg
W = 500N
F_{N} = Force normal to the plane
F = Frictional force (oposing the tendency of the block to move down the plane)
(i) W_{⊥} = W Cos 10°
W_{⊥} = 492N
(ii) W_{¶ }= W Sin 10°
W_{¶ }= 86.8N
(iii) System is in equilibrium so ΣF_{⊥} = 0
(W_{⊥}) + F_{N} = 0
F_{N} = 482N
(iv) System is in equilibrium so ΣF_{¶} = 0
(W_{¶}) + F = 0
F = 86.8N
(iv) The block will begin to slip when F = F_{max}
F_{N} = W_{⊥}
So F_{N} = W Cos θEq1
And F = W_{¶ }
So F = W Sin θ Eq2
But F = µW_{⊥}and W_{⊥} = F_{N}
F_{max} = µF_{N}
Dividing Eq2 by Eq1
(F_{max}) /
(F_{N}) = Tan θ
Since F_{max} =
Tan θ = (µF_{N}) /
(F_{N})
Tan θ = µ
θ = 16°
If θ = 16° then W Sin θ > µW Cos θ
When tan θ > µ the block will slip down the plane.








10. A farmer has a uniform bar which can swing out on a hinges to secure his access road. The 5m bar has a mass of 20kg and is supported by a cable hinged to the top of the gate post and attached to the middle of the bar. A second hinge is attached to the post at its mid point of the 2m post. Calculate
(a)
the tension, T, in the cable.
(b) the force on the hinge at the end of the bar.

Fig 10.1. 
Answer 
W = Weight of the bar
W = 200N
The bar is in equilibrium so:
ΣF_{BAR} = 0
Tan θ = ^{1}/_{2.5}
θ = 21.8 °
Vertical forces on the bar
ΣF_{v} = 0
[(W)] + T Cos θ = 0
T = ^{W}/_{Cos θ}
T = 215N
Horizontal forces on the bar
ΣF_{H} = 0
R + [(T Sn θ)] = 0
R = 80N






