Unit 3 / Practice Drills
Unit 3 Practice Drills
Friction and Trusses
A mixed unit: friction is direction-sensitive, while trusses are method-heavy and repetitive.
Unit 3 Practice Drills
Friction and trusses, organised by the repeated SPPU setups.
Drills 1 and 2 cover the repeated ladder-friction pattern. Drill 3 covers the wedge setup. Drills 4 and 5 use the same truss so students can compare the method of joints with the method of sections.
Drill 1 - Ladder Friction: Minimum Angle
Problem
A uniform ladder 5 m long and weighing 200 N leans against a smooth vertical wall. The coefficient of static friction between the ladder and the floor is μₛ = 0.4. Find the minimum angle the ladder makes with the floor for it to stay in equilibrium.
Figure 1A
Ladder against a smooth wall with floor friction
Q1 - Support reactions and force directions
There are three support reactions: N_A upward at the floor, F_A horizontally toward the wall, and N_B horizontally away from the smooth wall.
The wall is smooth, so no friction acts at B.
Q2 - Apply ΣFy = 0
N_A − W = 0, so N_A = W = 200 N.
Q3 - Apply ΣFx = 0
F_A − N_B = 0, so F_A = N_B.
Q4 - Take moments about A
N_B × (5 sinθ) − 200 × (2.5 cosθ) = 0
N_B = 100 cotθ
Q5 - Impending slip condition
At the minimum angle, F_A = μₛN_A = 0.4 × 200 = 80 N.
Since F_A = N_B, set 80 = 100 cotθ.
cotθ = 0.8, so tanθ = 1.25 and θ = 51.3°.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked ladder FBD with limiting reactions
Where students lose marks
The moment arm of the wall reaction is the vertical height 5 sinθ, while the moment arm of the ladder weight is the horizontal distance to its midpoint 2.5 cosθ. Students often use 5 and 2.5 directly instead of using the perpendicular distances.
Drill 2 - Ladder Friction: Person Climbing
Problem
A 6 m ladder weighing 120 N rests against a smooth wall at 65° to the horizontal. μₛ = 0.3 at the floor. A person weighing 650 N starts climbing from the bottom. How far up the ladder can the person go before the ladder slips?
Figure 1A
Person climbing a ladder at 65°
Q1 - FBD and unknowns
Known loads: 120 N at the ladder midpoint and 650 N at a distance x along the ladder from A.
Unknown reactions: N_A, F_A, and N_B.
Q2 - Find N_A and the limiting friction force
ΣFy = 0 gives N_A = 120 + 650 = 770 N.
At impending slip, F_A(max) = μₛN_A = 0.3 × 770 = 231 N.
That means N_B(max) = 231 N as well.
Q3 - Confirm ΣFx = 0
ΣFx = 0 gives F_A = N_B. At slip, both are 231 N.
Q4 - Moment equation about A
N_B(6 sin65°) − 120(3 cos65°) − 650(x cos65°) = 0
This is the correct moment balance because x is measured along the ladder, not horizontally.
Q5 - Solve for x
Substitute N_B = 231 N into the moment equation.
x = 4.02 m from the foot of the ladder.
That is the maximum distance the person can climb before slipping begins.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked ladder FBD with limiting climb distance
Where students lose marks
x is measured along the ladder, so the person’s moment arm about A is x cos65°, not x. Missing that cosine is the most common reason this answer comes out too small or too large.
Drill 3 - Wedge Friction
Problem
A 12° wedge is used to raise a block weighing 800 N. The coefficient of friction is 0.25 at all contact surfaces (wedge-floor and wedge-block). The block is constrained to move only vertically. Find the horizontal force P required to push the wedge.
Figure 1A
Separate FBDs for the block and the wedge
Q1 - Separate the two FBDs
At the wedge-block interface, N₁ and F₁ on the block are equal and opposite to the forces on the wedge.
Because the block rises vertically, friction on the block acts down the contact face.
Q2 - Block equilibrium to find N₁
Use α = 12° and F₁ = μN₁ = 0.25N₁.
For the block, ΣFy = 0 gives N₁cos12° − F₁sin12° − 800 = 0.
N₁(cos12° − 0.25 sin12°) = 800, so N₁ = 863.8 N.
Q3 - Find the friction force on the block
F₁ = μN₁ = 0.25 × 863.8 = 215.9 N.
The guide reaction balances the remaining horizontal component, but it is not needed in the final P calculation.
Q4 - Wedge vertical equilibrium to find N₂
On the wedge, the block pushes downward through N₁ and F₁.
ΣFy = 0 gives N₂ + F₁sin12° − N₁cos12° = 0.
So N₂ = 800 N.
Q5 - Wedge horizontal equilibrium to find P
The floor friction is F₂ = μN₂ = 0.25 × 800 = 200 N.
ΣFx = 0 gives P = N₁sin12° + F₁cos12° + F₂.
P = 863.8 sin12° + 215.9 cos12° + 200 = 590.7 N.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked wedge FBD with solved contact forces
Where students lose marks
There are two friction surfaces here, not one. F₁ acts at the wedge-block interface and F₂ acts at the wedge-floor interface. Missing either one gives a force P that is far too low.
Drill 4 - Truss: Method of Joints
Problem
A pin-jointed truss is supported by a pin at A and a roller at E. The bottom chord has joints A, B, C, D, E spaced 2 m apart. Joint F is 2 m above C. Members are AB, BC, CD, DE, AF, FE, BF, CF, and DF. Vertical loads are 15 kN at B and 25 kN at D. Find the force in every member and state tension or compression.
Figure 1A
Truss for the method of joints
Q1 - Support reactions
Taking moments about A: R_E × 8 − 15 × 2 − 25 × 6 = 0.
R_E = 22.5 kN upward.
Then ΣFy = 0 gives R_A = 17.5 kN upward, and ΣFx = 0 gives A_x = 0.
Q2 - Joint E
At joint E, only DE and FE are unknown.
F_DE = 45 kN tension.
F_FE = 50.3 kN compression.
Q3 - Joint D
Using the 25 kN load and the known force in DE:
F_CD = 20 kN tension.
F_DF = 35.4 kN tension.
Q4 - Joint C then joint B
Joint C is the smart next step because CD is known and CF is vertical.
At joint C: F_BC = 20 kN tension and F_CF = 0, so CF is a zero-force member.
At joint B: F_BF = 21.2 kN tension and F_AB = 35 kN tension.
Q5 - Verify at joint A and list all member forces
Using joint A with R_A = 17.5 kN gives F_AF = 39.1 kN compression.
Final member list:
AB = 35 T, BC = 20 T, CD = 20 T, DE = 45 T
AF = 39.1 C, FE = 50.3 C, BF = 21.2 T, CF = 0, DF = 35.4 T
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked truss with member-force summary
Where students lose marks
The fastest route is not E → D → F. Joint F still has too many unknowns at that stage. The clean route is E → D → C → B → A, where CF drops out as a zero-force member before you use the upper joint.
Drill 5 - Truss: Method of Sections
Problem
Use the same truss as Drill 4. Using the method of sections, find the forces in members AF, BF, and AB directly, without solving the full truss again.
Figure 1A
Broken section through AF, BF, and AB
Q1 - Reuse the support reactions
From Drill 4, R_A = 17.5 kN upward, R_E = 22.5 kN upward, and A_x = 0.
Q2 - Make the section cut
Use a broken section through AF, BF, and AB.
For the cleanest moment equations, take the A-side section after the cut.
Q3 - Take moments about joint B to find F_AF directly
Moments about B eliminate F_BF and F_AB.
Using the A-side section: R_A × 2 = F_AF × (2/√5).
F_AF = 39.1 kN compression.
Q4 - Take moments about joint F to find F_AB directly
Moments about F eliminate F_AF and F_BF.
R_A × 4 = F_AB × 2.
F_AB = 35 kN tension.
Q5 - Finish F_BF with one quick joint check
Go to joint B with the known 15 kN load and F_AB = 35 kN.
ΣFy = 0 gives F_BF sin45° = 15.
F_BF = 21.2 kN tension.
These match the method-of-joints results from Drill 4.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked section result for AF, BF, and AB
Where students lose marks
A single section cut gets AF and AB immediately because moments can eliminate the other cut forces. BF is then quickest from one joint check. Students often try to force all three from one section equilibrium set even when the geometry makes that inefficient.