Example 1 · Resolving by effect
Carpenters often drive a wedge in to lock a tenon joint. The right-triangle wedge shown has a base of 35 mm and a height of 5 mm. When it is driven in by a horizontal force F, the pressing force exerted by its slanted face is about ( )
A. 7F
B. 7.07F
C. 0.14F
D. 1.01F
Answer B Resolve F perpendicular to the slanted face and perpendicular to the base. The force triangle is similar to the wedge triangle:
Practice 1 · Multiple answers
As shown, a downward force F of 180 N is resolved into two components F₁ and F₂. The angle between F₁ and F is α = 37°, and the angle between F₂ and F is θ. Given sin 37° = 0.6 and cos 37° = 0.8, which statements are correct? ( )
A. When θ = 90°, F₂ = 240 N
B. When θ = 37°, F₂ = 112.5 N
C. When θ = 53°, F₂ = 144 N
D. Whatever the value of θ, F₂ can never be less than 108 N
Answer BD θ = 90°: F₂ = F tan 37° = 135 N; θ = 37°: F₁ = F₂ = 112.5 N; at θ = 53°, F₂ ⊥ F₁ and F₂ takes its minimum F sin 37° = 108 N
Model 1
Sliding Knots vs Fixed Knots
Sliding
Formed where a rope passes over a pulley or a smooth hook. It is really one rope: the tensions in the two segments are always equal, and their resultant lies along the angle bisector
Fixed
The rope on either side of the knot becomes two separate ropes, so the tensions in the two segments are not necessarily equal
A smooth hanger on a clothesline: the two tensions are always equal, and the angle depends only on the pole spacing and the rope length
Model 2
Hinged Rods vs Fixed Rods
Hinged rod: one end on a pivot or hinge
The force from a light hinged rod always acts along the rod
Fixed rod: one end built in (into a wall or the ground)
The force is not necessarily along the rod; find it from the equilibrium condition or Newton's second law
Example 2 · Sliding and fixed knots
As shown, a spider hangs at rest from a water pipe by its silk. Threads OM and ON make angles α and β with the vertical (α > β). Let F₁ and F₂ be the tensions in OM and ON. Then ( )
A. The vertical component of F₁ is greater than that of F₂
B. The vertical component of F₁ equals that of F₂
C. The horizontal component of F₁ is greater than that of F₂
D. The horizontal component of F₁ equals that of F₂
Answer D The threads meet in a fixed knot at O, so the two tensions need not be equal. At O, horizontally: F₁ sin α = F₂ sin β, so the horizontal components are equal; since α > β, F₁ has the smaller vertical component
Example 3 · Hinged and fixed rods
In figure (a), a light rope AD passes over a light pulley fixed to the right end of horizontal beam BC and holds a mass m₁, with ∠ACB = 30°. In figure (b), light rod HG is hinged to a vertical wall at one end and held by string EG at the other; EG makes 30° with the horizontal, and string GF hangs a mass m₂ from point G. With gravitational acceleration g, which statement is correct? ( )
A. In (a), the force from BC on the pulley is
B. In (b), the force on rod HG from the strings is
C. Tension ratio
:
=
D. Tension ratio
:
=
Example 3 Solution
(a) Fixed rod + pulley (sliding knot)
(b) Hinged rod + fixed knot
Answer D
; force from BC on the pulley is m₁g; force on rod HG is
Practice 2
A small crane: A and B are smooth light pulleys, C is a motor, and a light inextensible rope links them to object P. Pulley A's axle rides on a horizontal telescopic arm; pulley B rides on a vertical one. With P at rest ( )
A. The force on A's axle may be horizontal
B. The force on A's axle must exceed P's weight
C. Moving only A right increases the force on its axle
D. Moving only B up increases the force on A's axle
Answer C A is a sliding knot: both rope segments carry tension G, and the axle force is their resultant. The angle θ lies between 90° and 180°, so the resultant cannot be horizontal and can be greater than, equal to, or less than G. Moving A right makes θ smaller and the resultant larger; moving B up makes θ larger and the resultant smaller
Key ideas
2. Equilibrium of Concurrent Forces
1 Equilibrium
The object is at rest or in uniform straight-line motion, i.e. a = 0
2 Condition
Fnet = 0; equivalently, the net force in the x direction and in the y direction are each zero
3 Perpendicular components
Resolve every force along perpendicular x and y axes. Adding the x components gives Fₓ, adding the y components gives the net y force, and these combine into F
Choose axes so that as many forces as possible lie along them
Project each force onto the axes → add components head to tail on each axis → combine
Key ideas
3. Consequences of Equilibrium
2 forces
The two forces must be equal in size and opposite in direction
3 forces
Any one force is equal and opposite to the resultant of the other two
Placed head to tail, the three forces close a triangle
n forces
Any one force is equal and opposite to the resultant of all the others
From the textbook
A child slides down a slide inclined at θ at constant speed. If the angle θ is increased, do the forces on the child change? Can the child still slide down at constant speed?
Hint At constant speed, kinetic friction equals the component of gravity along the slope. With a steeper slope: gravity mg is unchanged; the normal force mg cos θ decreases, so friction μmg cos θ decreases; the component along the slope mg sin θ increases ⇒ no longer constant speed: the child speeds up
Check 1 · True or false
A stone arch has a top made of six identical stones, as sketched. Stones 1 and 6 are fixed; stones 2 and 5 each have mass m; stones 3 and 4 each have mass m′. Friction between stones is negligible, and gravitational acceleration is g.
(1) Stones 2, 3, 4 and 5 are each in equilibrium under three forces
(2) The normal force between stones 3 and 4 is zero
(3) The normal force of stone 1 on stone 2 equals (m + m′)g
(4) An object in equilibrium must have zero acceleration
True Each stone feels only gravity and the normal forces from its two contact faces
False The 3–4 contact face is vertical; without a normal force there, stone 3 could not balance
False Treating 2 and 3 as one system, only the vertical component of the force from 1 on 2 equals (m + m′)g
True Equilibrium means zero net force, so a = 0
Check 2
As shown, a football in a net bag hangs from point A
on a smooth wall, touching it at B. Its weight is G; rope AC
makes angle α with the wall. Wall force: F
N
; tension: F
T
.
Ignore the weight of the net. Which relation is correct? ( )
A. F
T
= F
N
B. F
T
< F
N
C. F
T
> G
D. F
T
< G
Answer C Gravity, the wall's normal force and the rope tension balance. Head to tail they form a right triangle with the tension as the hypotenuse: tension = G/cos α > G
Check 3
A 2 kg shot is placed on a 45° incline and held by a vertical board, so it is at rest. Ignore friction on the shot and take g = 10 m/s². Then ( )
A. The shot feels 4 forces: gravity, the normal force from the incline, the normal force from the board, and a "sliding-down force"
B. The force from the board on the shot points up the incline
C. The force of the shot on the board is
D. The force of the shot on the incline is
Answer D The shot feels only three forces: gravity G, the board's normal force F₁ (perpendicular to the board, horizontal to the right) and the incline's normal force F₂. A "sliding-down force" is a component of gravity, not an extra force
Topic 1 Free-body analysis
The picture shows a kitten leaping off the ground to chase a butterfly. Then ( )
A. The flying butterfly is acted on by gravity only
B. When the butterfly turns, the net force on it is along its direction of motion
C. In the air, the kitten is acted on by gravity and a normal force
D. While pushing off, the normal force on the kitten is greater than its weight
A ✗ The butterfly is also pushed by the air
B ✗ On a curved path the net force points toward the inside of the curve, not along the motion
C ✗ In the air the kitten touches nothing, so there is no normal force
D ✓ Pushing off, the kitten accelerates upward from rest, so the ground's normal force exceeds its weight
Answer D
Topic 1 Free-body analysis
After cleaning the board, Ming sticks a magnetic eraser onto the vertical blackboard and sees it slowly slide down by itself. He moves it to another spot, where it stays still. Which statement is correct? ( )
A. While slowly sliding down, the friction on the eraser is greater than its weight
B. While slowly sliding, the total force from the board on the eraser is vertically upward and equal to its weight
C. When at rest, the eraser feels only 3 forces
D. When at rest, the friction on the eraser is less than its weight
A ✗ Sliding slowly counts as constant speed, so it is in equilibrium and kinetic friction equals the weight
B ✓ The board's magnetic force, normal force and friction together balance gravity: vertically upward, equal to the weight
C ✗ At rest it feels 4 forces: gravity, magnetic force, normal force and static friction
D ✗ At rest, static friction balances gravity, so they are equal
Answer B
Example 1 · Composition and perpendicular components
A ball of mass m hangs from a vertical wall by string AB, which makes 60° with the wall. In figure (a), a pull F₁ at 120° to the string holds the ball at rest, and the string tension is Fa. In figure (b), a pull F₂ at 150° to the string holds the ball at rest, and the string tension is Fb. With gravitational acceleration g, which equation is correct? ( )
A.
B.
C.
F
a
D.
F
b
Answer A Composition: the resultant of the pull and the string tension is equal and opposite to mg. In (a) the two forces are symmetric about the vertical, so F₁ and the tension both equal mg. In (b):
string tension
Example 2 · Whole system and isolation
As shown, in one vertical plane, balls A and B are tied with inextensible strings a, b, c and d. The top of a hangs from a fixed vertical frame; d passes over the left fixed pulley and c over the right one, attached to identical counterweights P and Q. The pulley heights are adjusted until everything balances. Balls A, B and counterweights P, Q each have mass 50 g; strings c and d are parallel at θ = 30° to the horizontal (no friction). The tensions in a and b are (g = 10 m/s²) ( )
A. 2 N 1 N
B. 2 N 0.5 N
C. 1 N 1 N
D. 1 N 0.5 N
Example 2 Solution
Whole system for a, then isolate B for b
Step 1 Whole system → tension in a
Treat A and B as one system: the tensions in c and d both equal the counterweight's weight mg, are parallel and opposite, and cancel. So the tension in a = 2mg = 1 N
Step 2 Isolate B → tension in b
B feels 2mg upward from a, its weight mg, a pull mg from d (down-left at 30°), and the tension in b. Horizontal part mg cos 30°, vertical part 0.5mg; combining, the tension in b = mg = 0.5 N
Answer D
Practice · Whole system and isolation
As shown, an incline of angle α is fixed on level ground. Two uniform balls P and Q sit between the incline and a fixed vertical board; P has twice the mass of Q. All surfaces are smooth and the system is at rest. If the line joining the centres of P and Q makes angle β with the vertical, which is correct? ( )
A. 4 tan α = tan β
B. 3 tan α = tan β
C. 2 tan α = tan β
D. tan α = tan β
Answer B Let Q have mass m. Whole system: board force F₂ = 3mg tan α. Isolate Q: F₂ = mg tan β. Together: 3 tan α = tan β