Topic 2: Forces and Moments

Course: Prep4Uni Physics 1

Chapter 1: Quantities and Measurement
Chapter 2: Forces and Moments
Chapter 3: Motion and Forces
Chapter 4: Energy and Fields
Chapter 5: Projectile Motion

🚁Overview

This topic explores how objects interact through forces and how these forces produce both linear and rotational effects. Understanding forces and moments is essential for analyzing structures, machines, and motion in physics and engineering contexts.

📖Contents

Types of forces
Hooke’s Law and Elastic Forces
Moment and torque
Couples and the Principle of Moments
Equilibrium Conditions

🎯Learning Outcomes

By the end of this section, students should be able to:

Describe forces on masses, charges, and currents in various fields
Understand forces such as normal, frictional, buoyant, and viscous
Apply Hooke’s Law: F = kx
Define and apply moment of a force and torque
Understand the concept of a couple
Apply the principle of moments and define the centre of gravity
Analyze equilibrium using free-body diagrams and vector methods

🔢1. Types of Forces

Contact Forces:

Normal Force – Acts perpendicular to a surface (e.g., the table pushing up on a book).
Frictional Force – Acts opposite to the direction of motion (e.g., between tires and road).
Tension – Force transmitted through a string, rope, or cable.
Spring Force – Governed by Hooke’s Law: F = kx.

Non-Contact (Field) Forces:

Gravitational Force – Attraction between two masses: F = Gm₁m₂/r²
Electrostatic Force – Force between charges: F = kq₁q₂/r²
Magnetic Force – On moving charges or currents in magnetic fields.
Buoyant Force – Upward force on an object in a fluid.
Viscous Force – Resistance due to fluid motion (e.g., air drag, water resistance).

📐Learning-Check 1:
A 2 kg object is placed on a flat table. Identify all the forces acting on the object and state their directions.

Answer:

- Forces acting:
  - Weight (gravitational force) acting downward
  - Normal force from the table acting upward

📐Learning-Check 2:
Which of the following is a non-contact force?
A) Friction
B) Tension
C) Buoyant force
D) Gravitational force

Answer: D) Gravitational force
Explanation: Gravitational force acts at a distance without contact, unlike friction, tension, and buoyant force (which require a medium).

🔢2. Hooke’s Law and Elastic Forces

Hooke’s Law:

Describes the force exerted by a stretched or compressed spring:
F = kx,
where
F = force applied (N)
k = spring constant (N/m)
x = extension or compression (m)

Example:
A spring stretches 4 cm under a 2 N force. What is the spring constant?
Solution:
F = kx → k = F / x = 2 / 0.04 = 50 N/m

📐Learning-Check 1:
A spring extends 0.08 m when a 4 N force is applied. What is the spring constant?

Solution:
F = kx → k = F / x = 4 N / 0.08 m = 50 N/m

📐Learning-Check 2:
True or False: Hooke’s Law is valid for all amounts of deformation, even when the spring is permanently stretched.

Answer: False
Explanation: Hooke’s Law is only valid within the elastic limit of the spring. Beyond this, deformation may become permanent.

🔢3. Moment and Torque

Moment of a Force (Turning Effect):

Moment (M) = Force × Perpendicular Distance from Pivot
Unit: Newton-metre (Nm)

Torque is a type of moment causing rotation about an axis.

For rotation: Torque (τ) = r × F, where r is the lever arm vector, F is the force vector.

The picture below illustrate the moment of a force F acting a point point r from the support (end of the wall here).

A horizontal lever arm attached to a fixed wall. A downward force F is applied at the end of the lever, creating a torque about the pivot. The rotational effect (moment) M = F × r, where r is the lever arm length. — The rotational effect (moment) M = F × r, where F is the force and r is the lever arm length.

Example:
A 10 N force is applied perpendicularly at 0.5 m from a hinge.
Moment = 10 × 0.5 = 5 Nm

📐Learning-Check 1:
Calculate the moment produced by a 12 N force applied at a distance of 0.25 m from a pivot.

Solution:
Moment = Force × Distance = 12 N × 0.25 m = 3 Nm

📐Learning-Check 2:
A spanner applies a torque of 10 Nm. If the force applied is 20 N, what is the perpendicular distance from the axis of rotation?

Solution:
Torque = Force × Distance
10 Nm = 20 N × d → d = 10 / 20 = 0.5 m

🔢4. Couples and the Principle of Moments

Couple:

Two equal and opposite forces acting on different lines.
Produces pure rotation (no net force, only turning effect).

Principle of Moments:

For an object in rotational equilibrium:
Sum of clockwise moments = Sum of anticlockwise moments
The picture below illustrate the moment due to two equal forces F acting at distance D apart but in opposite direction forming a rotating moment called pure couple.

Example:
A uniform beam is balanced on a pivot. A 20 N weight is 1.5 m from the pivot on one side. What force must balance it 1 m from the pivot on the other side?
Clockwise Moment = 20 × 1.5 = 30 Nm
Let force F × 1 = 30 ⇒ F = 30 N

📐Learning-Check 1:
Two equal and opposite forces of 15 N act on a steering wheel, 0.4 m apart. What is the moment (torque) produced by this couple?

Solution:
Moment of a couple = One force × Distance between forces
= 15 N × 0.4 m = 6 Nm

📐Learning-Check 2:
A plank is balanced on a pivot. A 30 N weight is placed 1.2 m from the pivot. What weight must be placed 0.8 m on the other side to balance the plank?

Solution:
Clockwise moment = 30 N × 1.2 m = 36 Nm
To balance: Anticlockwise moment = W × 0.8 m → W = 36 / 0.8 = 45 N

🔢5. Equilibrium Conditions

Translational Equilibrium:

No net force → Object remains at rest or moves at constant velocity.
∑F_x = 0, ∑F_y = 0

Rotational Equilibrium:

No net moment → Object does not rotate or rotates at constant rate.
∑Moments = 0

Free-Body Diagrams (FBDs):

Diagrams showing all forces acting on a body. Essential for analyzing equilibrium.
Below is a free-body diagram of a block in static equilibrium on a horizontal surface, showing the pulling force (F), frictional force (f), weight (mg), and normal reaction (N).

Free Body Diagram showing the Forces acting on an object sitting on a flat table

Below is a free-body diagram of a block on a plane inclined at θº to the horizontal. The weight of the body is decomposed into one that is along the plane and the other perpendicular to the plane.

Free-body Diagram for forces acting on an object on an inclined plane.

Beam Equilibrium

A beam is said to be in equilibrium when there is no tendency for it to move. There are two conditions for equilibrium, namely, the sum of the forces acting vertically downward must be equal to the sum of the forces acting vertically upward and the total moment of the forces acting on a beam must be zero. That is, there are both translational equilibrium and rotational equilibrium.

Example:
A ladder leans against a wall. Drawing the FBD helps identify all forces: weight, normal reactions, friction, etc., and allows equilibrium equations to be written and solved.

📐Learning-Check 1:
Draw a free-body diagram for a box being pushed across a rough floor at constant speed. Label all forces.

Answer (labeled FBD):

- Weight acting downward
- Normal force from the floor acting upward
- Applied force pushing the box forward
- Frictional force acting opposite to motion (backward)

📐Learning-Check 2:
A beam is in static equilibrium. Which of the following must be true?
A) ∑F ≠ 0 and ∑M = 0
B) ∑F = 0 and ∑M = 0
C) ∑F = 0 and ∑M ≠ 0
D) ∑F ≠ 0 and ∑M ≠ 0

Answer: B) ∑F = 0 and ∑M = 0
Explanation: For an object to be in complete equilibrium (static or constant velocity), both the net force and net moment must be zero.

💡Some Interesting Examples

Example 1: Equilibrium of Forces: Vector‐Method Example

Problem: Two forces act concurrently at point O:

A = 80 N directed horizontally to the right
B = 60 N directed vertically upward

Find the single reaction force R (magnitude and direction) which, when applied at O, will keep the point in equilibrium (∑ F = 0).

Section 1: Graphical Vector‐Addition Method

At O, draw vector A to the right (length ∝ 80 N).
From the tip of A, draw vector B upward (length ∝ 60 N).
The closing vector from the tip of B back to O is the reaction R, of equal magnitude but opposite direction to A + B.

Note that the magnitude and direction of the resultant vector can be obtained by measurement of the length and the angle respectively.

Section 2: Resolved‐Component Method

1. Resolve each force into Cartesian components:

A =（80，0），B = （0，60）

A+B = (80+0, 0+60) = (80, 60)

2. Sum the components to get the resultant that must be opposed:

A+B = (80+0, 0+60) = (80, 60)  
R = - (A+B) = - (80, 60) = (-80, -60)

3. Compute the magnitude of R:

 | R |= √[(–80)² + (–60)²] = √(6400 + 3600) = 100 N

4. Compute the direction of R (θ measured from the positive x-axis):

  θ = atan2(–60, –80)  
    = 180° + atan(60/80)  
    ≈ 180° + 36.87°  
    = 216.87°

So | R | = 100 N acting at 216.9° from the +x-axis (i.e. 36.9° below the –x direction).

Example 2: Centre of Gravity of Three Discrete Bodies (Sun–Earth–Moon)

Problem. Three astronomical bodies lie approximately on a straight line:

The Sun, mass M_☉ = 1.989 × 10³⁰ kg, at x = 0.
The Earth, mass M_E = 5.972 × 10²⁴ kg, at x = 1 AU = 1.496 × 10¹¹ m.
The Moon, mass M_m = 7.347 × 10²² kg, at x = 1.496 × 10¹¹ + 3.84 × 10⁸ = 1.49984 × 10¹¹ m.

Solution.

List masses and positions:
M_☉ = 1.989 × 10³⁰ kg, x_☉ = 0 M_E = 5.972 × 10²⁴ kg, x_E = 1.496 × 10¹¹ m M_m = 7.347 × 10²² kg, x_m = 1.49984 × 10¹¹ m
Total mass:
M_tot = M_☉ + M_E + M_m ≈ 1.989005972 × 10³⁰ kg

Weighted sum of positions:

Σ M_i x_i = 
   M_☉·x_☉ + M_E·x_E + M_m·x_m
 ≈ 0 + (5.972×10⁴⁴) + (7.347×10³³)
 ≈ 9.039×10³⁵ kg·m

Centre of gravity:

x_CG = (Σ M_i x_i) / M_tot
             ≈ (9.039×10³⁵ kg·m) / (1.989005972×10³⁰ kg)
             ≈ 4.546×10⁵ m

CG of Sun, Earth and Moon (not to scale)

Answer.
The Sun–Earth–Moon system’s centre of gravity lies at
x_CG ≈ 4.55 × 10⁵ m
from the Sun’s centre (about 455 km along the Sun–Earth line, well inside the Sun’s radius of ~696,000 km).

Example 3: Maximum Incline Before Sliding

Problem.
A box of mass m = 10 kg rests on a rough incline. The coefficient of static friction between the box and the plane is μ = 0.20.
What is the maximum angle θ_max (with the horizontal) at which the plane can be inclined without the box beginning to slide?

Free-Body Diagram

Solution

Forces along the plane:

F_g,‖ = mg sinθ
f_max = μ N = μ mg cosθ

Equilibrium at the threshold of sliding:

F_g,‖ = f_max
mg sinθ_max = μ mg cosθ_max

Solve for θ_max:
Divide both sides by mg cosθ_max:

tan θ_max = μ
θ_max = atan(0.20) ≈ 11.31°

Answer.
The maximum incline angle is
θ_max ≈ 11.3°.

⚙️Key Concepts Recap

Concept	Formula	Notes
Hooke’s Law	F = kx	Linear spring force
Moment	M = F × d	d = perpendicular distance to pivot
Torque	τ = r × F	Vector form
Equilibrium	∑F = 0, ∑M = 0	No acceleration or rotation

Proceed to: Topic 3: Motion and Forces

Go back to Prep4Uni Physics 1

📝 EXERCISES

Forces and Moments: Questions & Answers

1. Types of Forces

Q1: A block rests on an inclined plane at 30°. List all the forces acting on it, describe their directions, and explain how each arises.

A1:

- Weight (mg), acting vertically downward due to gravity.
- Normal force (N), acting perpendicular to the plane, from contact with the surface.
- Frictional force (f), acting parallel to the plane, opposing any tendency to slide (static if at rest, kinetic if sliding).
- (If tied) Tension (T), along the rope if the block is attached.

Q2: Explain the difference between viscous drag and buoyant force, including their origins and how they scale with velocity or volume.

A2:

- Viscous drag arises from fluid viscosity; for a sphere in laminar flow F_d = 6π η r v (Stokes’ law), scaling linearly with speed v.
- Buoyant force arises from pressure differences in a fluid; Archimedes’ principle: F_b = ρ_fluid V g, scaling with the displaced volume V, independent of velocity.

Q3: A small metal bead is suspended from a thread in Earth’s gravitational field and then charged and placed in a uniform electric field. Identify and describe the forces acting on the bead in each scenario.

A3:

- Gravitational only: Weight mg downward and tension T upward in the thread.
- With electric field: Adds electrostatic force F_e = qE (upward if field opposes weight, downward if it augments), altering the thread tension accordingly.

Q4: Compare contact versus non-contact forces by giving two distinct engineering examples where each type is dominant.

A4:

- Contact: Brakes on a car (frictional force between pads and rotor); Gears in a gearbox (normal forces between teeth).
- Non-contact: Lift on an aircraft wing (pressure-field force); Magnetic levitation trains (magnetic repulsion).

Q5: Derive the expression for the net force on a body moving horizontally through a viscous fluid when both drag and an applied force act on it, and state the condition for terminal velocity.

A5:

- Net force: F_net = F_applied − F_drag.
- For laminar drag, F_drag = 6π η r v, so m a = F_applied − 6π η r v.
- Terminal velocity when a = 0: v_t = F_applied / (6π η r).

2. Hooke’s Law and Elastic Forces

Q6: A spring of natural length 0.20 m is stretched to 0.24 m by a 10 N force. What assumptions underlie Hooke’s Law, and how would the behavior change beyond the elastic limit?

A6:

- Assumptions: Linear response, no plastic deformation, constant spring constant k.
- Beyond elastic limit: Material yields, curve becomes non-linear, permanent deformation occurs; F no longer ∝ x.

Q7: Two springs in series each have constant k. Derive the equivalent spring constant k_s for the pair.

A7:

- Same force F stretches each by F/k, so total extension x = 2F/k.
- Equivalent k_s satisfies F = k_s x = k_s(2F/k), hence k_s = k/2.

Q8: Two springs in parallel each with constant k support a mass. Derive the combined constant k_p.

A8:

- Same extension x, each carries load: total force F = kx + kx = 2kx.
- Equivalent constant k_p = 2k.

Q9: A spring–mass system oscillates with small amplitude. Explain qualitatively how Hooke’s Law underpins simple harmonic motion and derive the angular frequency ω.

A9:

- Restoring force F = −k x ⇒ equation m ẍ + k x = 0.
- Solution is sinusoidal with ω² = k/m ⇒ ω = √(k/m).

Q10: A spring stretched by 0.05 m stores potential energy U = ½ k x². Show that half the work done in stretching goes into spring energy and half into kinetic energy if the mass attached is released from rest.

A10:

- Work done by external agent: W = ∫₀ˣ k x′ dx′ = ½ k x².
- If released, force is conservative; at equilibrium point all ½ k x² becomes kinetic, illustrating energy exchange.

3. Moment and Torque

Q11: Define the moment of a force about a point and explain why only the perpendicular lever arm enters the calculation.

A11:

- M = F d_⊥, where d_⊥ is the shortest distance from pivot to the force line.
- Only the perpendicular component produces turning effect; any component through the pivot produces zero moment.

Q12: A wrench applies a force at an angle. Show how to compute the effective lever arm and the moment produced.

A12:

- Perpendicular component F_⊥ = F sin θ.
- Moment M = F_⊥ r = F r sin θ.

Q13: Explain, with a sketch, why two equal and opposite forces offset by distance d produce a pure couple and no net force.

A13:

- Forces cancel vectorially, but their lines of action are separated by d, yielding net torque τ = F d.
- Sketch: two parallel arrows equal magnitude, opposite direction, separated by distance.

Q14: In the vector form τ = r × F, explain the significance of the right-hand rule and how it determines the axis of rotation.

A14:

- Cross product gives a vector perpendicular to the plane of r and F; right-hand rule (thumb) shows direction of the torque vector (axis).

Q15: A door 0.8 m wide needs 40 N applied perpendicular at its handle to open with 32 Nm torque. If you apply at 45° instead, what force is required?

A15:

- Effective lever arm: r sin 45° = 0.8·0.707 = 0.566 m.
- Required force: F = τ / (r sin 45°) = 32 / 0.566 ≈ 56.6 N.

4. Couples and the Principle of Moments

Q16: State the principle of moments and apply it to a seesaw with unequal child masses to find the balance point.

A16:

- Sum of clockwise moments = sum of anticlockwise moments.
- If child A (mass m₁) sits d₁ from pivot and child B (mass m₂) sits d₂, equilibrium when m₁g d₁ = m₂g d₂.

Q17: A uniform beam of weight W is supported at its ends. Show how to find the reaction forces at each support.

A17:

- Moment about one end: R₂ L = W (L/2) ⇒ R₂ = W/2.
- By symmetry, R₁ = W/2.

Q18: Two equal and opposite 25 N forces are applied 0.4 m apart. Calculate the couple’s moment and explain why it does not depend on the point of reference.

A18:

- Moment M = 25 N × 0.4 m = 10 Nm.
- Couple is a free vector; its torque is the same about any point since net force is zero.

Q19: On a rigid body in equilibrium under three non-parallel coplanar forces, explain why the three lines of action must be concurrent.

A19:

- If non-concurrent, their moments about some point cannot all cancel unless there’s a couple—contradicting pure force equilibrium.

Q20: A beam has a downward 100 N load at 1.2 m from pivot and an upward reaction of 60 N at 0.5 m on the other side. What additional force must act to satisfy the principle of moments?

A20:

- Clockwise moment = 100 N × 1.2 m = 120 Nm.
- Upward 60 N at 0.5 m gives anticlockwise 30 Nm.
- Additional anticlockwise moment 90 Nm needed ⇒ 90 N upward at 1 m from pivot.

5. Equilibrium Conditions

Q21: Write down the two conditions for complete static equilibrium of a rigid body in a plane and explain their physical meaning.

A21:

- ΣF_x = 0, ΣF_y = 0 ⇒ no net translation.
- ΣM = 0 (about any point) ⇒ no net rotation.

Q22: Sketch a free-body diagram for a block being pulled at constant velocity across a rough surface and derive the condition on forces.

A22:

- Forces: Applied F forward, friction f backward, weight mg down, normal N up.
- Equilibrium ⇒ F = f = μ N, N = mg.

Q23: A ladder of length L leans at angle θ against a smooth wall. List all forces, write equilibrium equations, and comment on the role of friction.

A23:

- At base: normal N₁ up, friction F_f horizontal. At top: normal N₂ horizontal. Weight mg at midpoint.
- ΣF_x: F_f − N₂ = 0.
- ΣF_y: N₁ − mg = 0.
- ΣM_base: N₂ L sinθ − mg (L/2 cosθ) = 0. Friction must be ≥ N₂ to prevent slipping.

Q24: Explain why it’s sufficient to sum moments about just one point when solving for equilibrium of a rigid body.

A24:

- With ΣF = 0, forces are balanced; summing moments about one pivot eliminates unknown at that point, giving a solvable equation.

Q25: A uniform plank is supported at its ends and carries two asymmetric loads. Describe your strategy—using free-body diagrams and equilibrium equations—to find both support reactions.

A25:

- Draw FBD with reactions R₁, R₂ and weights at known positions.
- Use ΣM about support 1 to find R₂, then ΣF_y = 0 to find R₁.

Forces and Moments: Problems & Solutions

1. Types of Forces

Problem 1: A 5 kg block lies on a horizontal table. List all forces acting on it and compute the normal force.

Solution:
Forces: weight (mg = 5×9.81 = 49.05 N down), normal force N upward. Equilibrium ⇒ N = 49.05 N.

Problem 2: A sphere of radius 0.1 m moves at 0.5 m/s through oil (η = 0.2 Pa·s). Calculate the viscous drag.

Solution:
Stokes’ law: F_d = 6π η r v = 6π×0.2×0.1×0.5 ≈ 0.188 N.

Problem 3: A 2 kg bead on a string is in equilibrium in a downward electric field of 400 N/C. What charge must the bead carry?

Solution:
Equilibrium: qE = mg ⇒ q = (2×9.81)/400 ≈ 0.0491 C.

Problem 4: Compare the magnitudes of the gravitational force and electrostatic force between two protons separated by 1 fm.

Solution:
F_g ≈ 1.86×10⁻³⁵ N; F_e ≈ 2.30×10⁻⁸ N.

Problem 5: A body experiences an applied force of 10 N and a drag force F_d = 2v (in N). Determine the terminal velocity.

Solution:
At terminal velocity: 10 = 2v ⇒ v = 5 m/s.

2. Hooke’s Law and Elastic Forces

Problem 6: A spring extends by 0.04 m under a 2 N load. Find the spring constant k.

Solution:
k = F/x = 2 / 0.04 = 50 N/m.

Problem 7: Two springs (k₁ = 100 N/m, k₂ = 200 N/m) are connected in series. Find the equivalent k.

Solution:
1/kₛ = 1/100 + 1/200 = 0.015 ⇒ kₛ ≈ 66.7 N/m.

Problem 8: A mass m = 0.5 kg on a spring (k = 80 N/m) is released from rest at x = 0.1 m. Find maximum speed.

Solution:
½k x² = ½m v² ⇒ v = x√(k/m) = 0.1√(80/0.5) ≈ 1.26 m/s.

Problem 9: Two identical springs (k = 120 N/m) in parallel support a 240 N weight. Find the extension.

Solution:
kₚ = 2k = 240 N/m ⇒ x = F/kₚ = 240/240 = 1.0 m.

Problem 10: A mass-spring system oscillates with period T = 0.5 s and m = 0.2 kg. Find k.

Solution:
T = 2π√(m/k) ⇒ k = 4π²m/T² ≈ 31.6 N/m.

3. Moment and Torque

Problem 11: A 30 N force is applied perpendicular to a lever arm 0.3 m from the pivot. Calculate the moment.

Solution:
M = F·d = 30×0.3 = 9 N·m.

Problem 12: A wrench 0.25 m long is used at 60°. What moment results from a 40 N force?

Solution:
M = F·r·sinθ = 40×0.25×sin60° ≈ 8.66 N·m.

Problem 13: A door 0.8 m wide needs 20 N at 30° to open with torque 12 N·m. Verify whether 20 N is sufficient.

Solution:
Effective arm = 0.8 sin30° = 0.4 m ⇒ M = 20×0.4 = 8 N·m (insufficient).

Problem 14: A torque of 15 N·m is required. What force at 0.2 m perpendicular arm?

Solution:
F = M/d = 15/0.2 = 75 N.

Problem 15: Two forces of 25 N produce a net moment of 7.5 N·m on opposite sides of a pivot. Find their separation.

Solution:
d = M/F = 7.5/25 = 0.3 m.

4. Couples and the Principle of Moments

Problem 16: Two 15 N forces act opposite, 0.4 m apart. Calculate the couple’s moment.

Solution:
τ = F·d = 15×0.4 = 6 N·m.

Problem 17: A seesaw of length 4 m carries a 30 kg child at 1 m from pivot. What mass balances at 1.5 m?

Solution:
m₂ = (30×1)/1.5 = 20 kg.

Problem 18: A uniform beam weight 100 N is pivoted at center. Two opposite 60 N forces act at ends. Determine net moment.

Solution:
Net force = 0; couple moment = 60×(L/2)+60×(L/2) = 60L (pure couple).

Problem 19: Show with numbers why a couple’s moment is independent of reference point.

Solution:
Changing pivot shifts lever arms equally for both forces; net M = F·d remains unchanged.

Problem 20: A wrench applies +20 N and –20 N at ends 0.5 m apart. Calculate total moment about midpoint.

Solution:
τ = 20×0.5 = 10 N·m.

5. Equilibrium Conditions

Problem 21: A block is pulled at constant speed by 50 N on a rough floor (μ=0.3). Find frictional force.

Solution:
N = mg = 49.05 N; f = μN = 0.3×49.05 = 14.72 N (so applied force must balance 14.72 N to maintain constant speed).

Problem 22: A ladder 5 m long leans at 60°; weight 600 N at midpoint. Find minimum μ at base to prevent slipping.

Solution:
R_w = (600×2.5×cos60°)/(5×sin60°) ≈ 173.2 N; N = 600 N; μ = 173.2/600 ≈ 0.29.

Problem 23: A simply supported beam 6 m long carries 800 N at 2 m from A. Find reactions at A and B.

Solution:
R_B·6 = 800·2 ⇒ R_B = 266.7 N; R_A = 800 − 266.7 = 533.3 N.

Problem 24: A plank 3 m long, weight 200 N, has loads 150 N at 0.5 m and 100 N at 2 m from left support. Find support forces.

Solution:
ΣM_A: R_B·3 = 150·0.5 + 100·2 = 75 + 200 = 275 ⇒ R_B = 91.7 N; R_A = 200+150+100 − 91.7 = 358.3 N.

Problem 25: A beam hinged at A and roller at B supports loads 300 N at 1 m and 200 N at 3 m along a 5 m beam. Find reactions at A and B.

Solution:
ΣM_A: R_B·5 = 300·1 + 200·3 = 300 + 600 = 900 ⇒ R_B = 180 N; ΣF_y: R_A + R_B = 500 ⇒ R_A = 320 N.

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