The Lamp on Two Strings
A lamp of weight $W$ hangs at rest from two light strings, each making an angle $\theta$ with the ceiling. The lamp does not move, so the two tensions together must exactly cancel the weight. Set $W$ and $\theta$ and watch how steeply the tensions grow as the strings flatten out.
each tension $T = \dfrac{W}{2\sin\theta}$ = 23.3 N · check $2T\sin\theta$ = 30.0 N
Resolving vertically gives $2T\sin\theta = W$. The two horizontal components $T\cos\theta$ are equal and opposite, so they cancel.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A body is in equilibrium. The resultant force on it is:
R2. The vertical component of a force $F$ at angle $\theta$ to the horizontal is:
R3. Resolving is useful because components:
A body is in equilibrium when the resultant of all the forces acting on it is zero. Resolving along any two perpendicular directions turns that single statement into two equations:
A body moving with constant velocity is also in equilibrium — the resultant is still zero, it is simply not at rest.
Show every force acting on the chosen body, each with a name and, where known, a value.
Usually horizontal and vertical, or parallel and perpendicular to a slope. State which way is positive.
A force at angle $\theta$ to the positive direction contributes $F\cos\theta$ along it and $F\sin\theta$ perpendicular to it.
Two equations, usually two unknowns. Solve them simultaneously.
If exactly three forces keep a body in equilibrium, drawing the three forces head-to-tail closes a triangle. The triangle can then be solved with trigonometry, which is often quicker than resolving.
Worked example: a lamp of weight $30\ \text{N}$ on two strings at $40^\circ$ to the horizontal has $2T\sin 40^\circ = 30$, so $T = 23.3\ \text{N}$ in each string.
Groups of 4: tap a card to pick it up, then tap a zone to file it — "✓ True for equilibrium" or "✗ False / a slip". Discuss each claim before filing it, then press Check cards. Six claims about resultant forces, resolved components and hanging bodies.
Two or three known forces are drawn from a point. Use the sliders to supply the magnitude F and direction theta of the force that makes everything balance. The dashed red resultant shrinks towards zero as you home in, and the panel confirms equilibrium. Then answer the challenge.
A lamp of weight $24\ \text{N}$ hangs at rest from two light strings, each making an angle of $50^\circ$ with the horizontal. Find the tension in each string.
The diagram is symmetric, so the two tensions are equal. Resolve vertically: each tension has a vertical component $T\sin 50^\circ$.
Each string has horizontal component $T\cos 50^\circ$, one to the left and one to the right. These cancel, so the horizontal equation tells us the tensions are equal.
The two vertical components must carry the weight: $2T\sin 50^\circ = 24$.
Resolving one tension: the vertical parts $T\sin 50^\circ$ add to $24\ \text{N}$, while the horizontal parts $T\cos 50^\circ$ are equal and opposite.
$T = \dfrac{24}{2\sin 50^\circ} = \dfrac{24}{1.532} = 15.7\ \text{N}$.
A particle is in equilibrium under three forces. Two of them are $6\ \text{N}$ due east and $8\ \text{N}$ due north. Find the magnitude and direction of the third force.
Find the resultant of the two known forces first. The third force must be equal and opposite to it.
$R = \sqrt{6^{2}+8^{2}} = 10\ \text{N}$ at $\theta = \tan^{-1}\!\left(\dfrac{8}{6}\right) = 53.1^\circ$ north of east.
The third force must be $10\ \text{N}$ in the opposite direction, that is $53.1^\circ$ south of west.
The resultant $R$ of the two known forces is opposed exactly by the third force, so the three forces balance.
East: $6 - 10\cos 53.1^\circ = 6 - 6 = 0$. North: $8 - 10\sin 53.1^\circ = 8 - 8 = 0$. Both directions balance.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A body is in equilibrium under several forces. Which statement must be true?
Q2. A weight of $50\ \text{N}$ hangs from two light strings, each making $30^\circ$ with the horizontal. The tension in each string is:
Q3. Three forces keep a particle in equilibrium. Drawn head-to-tail, the three forces:
Q4. A weight of $40\ \text{N}$ hangs from two light strings, each making $60^\circ$ with the horizontal. The tension in each string is:
Work each on paper, then reveal the solution to check your working.
A sign of weight $20\ \text{N}$ hangs at rest from two light strings, each making an angle of $35^\circ$ with the horizontal. Find the tension in each string.
Reveal Solution
Resolve vertically: $2T\sin 35^\circ = 20$, so $T = \dfrac{20}{2\sin 35^\circ} = 17.4\ \text{N}$ in each string.
A particle is in equilibrium under three forces. Two of them are $5\ \text{N}$ due east and $12\ \text{N}$ due north. Find the magnitude and direction of the third force.
Reveal Solution
The two known forces have resultant $\sqrt{5^{2}+12^{2}} = 13\ \text{N}$ at $\tan^{-1}(12/5) = 67.4^\circ$ north of east. The third force is $13\ \text{N}$ in the opposite direction, i.e. $67.4^\circ$ south of west.
A load of weight $100\ \text{N}$ hangs from a rope. A horizontal force $P$ pulls the load aside until the rope makes an angle of $30^\circ$ with the vertical. Find $P$ and the tension in the rope.
Reveal Solution
Resolve vertically: $T\cos 30^\circ = 100$, so $T = 115.5\ \text{N}$. Resolve horizontally: $P = T\sin 30^\circ = 57.7\ \text{N}$.
A particle of weight $50\ \text{N}$ is held at rest by two light strings, one making $30^\circ$ with the horizontal and the other $60^\circ$ with the horizontal, on opposite sides. Find the tension in each string.
Reveal Solution
Let the tensions be $T_{1}$ (at $30^\circ$) and $T_{2}$ (at $60^\circ$). Horizontally $T_{2}\cos 60^\circ = T_{1}\cos 30^\circ$, so $T_{2} = \sqrt{3}\,T_{1}$. Vertically $T_{1}\sin 30^\circ + T_{2}\sin 60^\circ = 50$, giving $0.5T_{1} + 1.5T_{1} = 50$. Hence $T_{1} = 25\ \text{N}$ and $T_{2} = 43.3\ \text{N}$.
Three forces act on a particle in equilibrium: $5\ \text{N}$, $12\ \text{N}$ and $13\ \text{N}$. The $5\ \text{N}$ and $12\ \text{N}$ forces are perpendicular. Find the angle between the $13\ \text{N}$ force and the $12\ \text{N}$ force.
Reveal Solution
The resultant of the $5\ \text{N}$ and $12\ \text{N}$ forces is $\sqrt{5^{2}+12^{2}} = 13\ \text{N}$, making $\tan^{-1}(5/12) = 22.6^\circ$ with the $12\ \text{N}$ force. The $13\ \text{N}$ force must be opposite to that resultant, so the angle between it and the $12\ \text{N}$ force is $180^\circ - 22.6^\circ = 157.4^\circ$.
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