Activity 2.1 4 Calculating Moments Answer Key – a thorough guide that walks learners through the fundamentals of torque, the step‑by‑step methodology for solving moment problems, and the correct solutions for the designated activity. This article is crafted for high‑school physics students, exam‑preparers, and anyone seeking a clear, structured approach to mastering moments in rotational dynamics Worth keeping that in mind..
Understanding the Concept of Moments
A moment (also called torque) quantifies the tendency of a force to rotate an object about a pivot point, known as the fulcrum. The magnitude of a moment depends on two variables: the magnitude of the applied force (F) and the perpendicular distance from the line of action of the force to the fulcrum (d). The basic relationship is expressed as:
[ \text{Moment (Nm)} = F \times d ]
Key points to remember
- Direction matters: A clockwise rotation is often considered negative, while counter‑clockwise is positive, depending on the chosen sign convention.
- Units: The International System of Units (SI) uses newtons (N) for force and meters (m) for distance, yielding moments in newton‑meters (Nm).
- Line of action: Only the component of the force that is perpendicular to the lever arm contributes to the moment; any parallel component produces no turning effect.
Key Formulae and Definitions
| Symbol | Meaning | Unit |
|---|---|---|
| F | Force applied | newtons (N) |
| d | Perpendicular distance from fulcrum to line of action of F | meters (m) |
| τ | Torque or moment | newton‑meters (Nm) |
| r | Lever arm (distance from fulcrum to point of force application) | meters (m) |
When multiple forces act on a rigid body, the net moment about a point is the algebraic sum of the individual moments:
[ \tau_{\text{net}} = \sum_{i} (F_i \times d_i) ]
If the net moment equals zero, the body is in rotational equilibrium (no angular acceleration) It's one of those things that adds up..
Step‑by‑Step Procedure for Solving Problems
- Identify the fulcrum – Locate the pivot point about which moments are taken.
- Draw a free‑body diagram – Represent all forces acting on the object, labeling magnitudes and points of application.
- Determine the lever arm – Measure the perpendicular distance from the fulcrum to each force’s line of action.
- Calculate each moment – Multiply the force by its lever arm; assign a sign (+ for counter‑clockwise, – for clockwise).
- Sum the moments – Add all signed moments to obtain the net moment.
- Apply equilibrium conditions (if required) – Set the net moment to zero for static equilibrium or use Newton’s second law for rotation ((\tau = I\alpha)) for dynamic scenarios.
- Solve for the unknown – Rearrange the equation to isolate the desired variable (force, distance, or acceleration).
Tip: Always double‑check units; converting centimeters to meters early prevents arithmetic errors Easy to understand, harder to ignore..
Worked Example 1
A uniform beam 4 m long is hinged at one end and held horizontally by a rope attached 1 m from the hinge. A 10 N weight hangs from the free end That's the part that actually makes a difference. Worth knowing..
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Fulcrum: hinge at the left end Simple, but easy to overlook..
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Forces:
- Weight: (F_w = 10 N) acting downward at 4 m from hinge → lever arm (d_w = 4 m).
- Rope tension: unknown (T) acting upward at 1 m from hinge → lever arm (d_T = 1 m).
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Moments about hinge:
- Weight moment (clockwise): (\tau_w = 10 N \times 4 m = 40 Nm) (negative).
- Rope moment (counter‑clockwise): (\tau_T = T \times 1 m) (positive).
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Set net moment to zero (static equilibrium): [ \tau_T - \tau_w = 0 ;\Rightarrow; T \times 1 m - 40 Nm = 0 ]
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Solve: (T = 40 N). Thus, the rope must exert a tension of 40 N to keep the beam horizontal.
Worked Example 2
A seesaw is 6 m long, with a 15 kg child sitting 2 m from the centre. Where must a 20 kg child sit on the opposite side to balance the seesaw?
- Convert masses to forces: (F_1 = 15 kg \times 9.81 m/s^2 = 147.15 N).
- Lever arm for first child: (d_1 = 2 m).
- Moment from first child: (\tau_1 = 147.15 N \times 2 m = 294.3 Nm) (clockwise).
- Force of second child: (F_2 = 20 kg \times 9.81 m/s^2 = 196.2 N).
- Let (d_2) be the unknown distance from the centre on the opposite side.
- Set moments equal for balance:
[ \tau_2 = \tau_1 ;\Rightarrow; 196.2 N \times d_2 = 294.3 Nm]
- Solve for (d_2):
[ d_2 = \frac{294.
To balance the seesaw, the moments about the fulcrum (center) must be equal. 15 , \text{N} \times 2 , \text{m} = 294.On the flip side, 3 , \text{Nm}) gives (d_2 = 1. 2 , \text{N} \times d_2 = 294.Solving (196.The 20 kg child must generate an equal counter-clockwise moment. The 15 kg child creates a clockwise moment of (147.3 , \text{Nm}). 5 , \text{m}) Small thing, real impact..
Conclusion: The 20 kg child must sit 1.5 meters from the center on the opposite side to achieve equilibrium And that's really what it comes down to. Less friction, more output..
