Standard +0.3 This is a standard three-force equilibrium problem requiring resolution of forces in two perpendicular directions and basic trigonometry. While it involves multiple particles and strings over pulleys, the solution follows a routine method (tensions equal weights, resolve at X, solve simultaneous equations) that is typical for M1 mechanics with no novel insight required.
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\includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-2_606_843_1731_651}
The diagram shows three particles \(A , B\) and \(C\) hanging freely in equilibrium, each being attached to the end of a string. The other ends of the three strings are tied together and are at the point \(X\). The strings carrying \(A\) and \(C\) pass over smooth fixed horizontal pegs \(P _ { 1 }\) and \(P _ { 2 }\) respectively. The weights of \(A , B\) and \(C\) are \(5.5 \mathrm {~N} , 7.3 \mathrm {~N}\) and \(W \mathrm {~N}\) respectively, and the angle \(P _ { 1 } X P _ { 2 }\) is a right angle. Find the angle \(A P _ { 1 } X\) and the value of \(W\).
For using triangle of forces or for resolving in dir° XP₁ or for using Lami's theorem or for resolving forces at X vertically and horizontally (equations must contain not more than one unknown angle)
M1
For correct \(\Delta\) or resolve XP₁ and \(\cos\alpha = \frac{5.5}{7.3}\); or \(5.5/\sin(90° + \alpha) = 7.3/\sin90°\) (Lami); or \(5.5\cos\alpha + W\sin\alpha = 7.3\) and \(5.5\sin\alpha = W\cos\alpha\)
A1
Angle AP₁X = 41.1° or 0.718°
A1
For correct triangle and \(W^2 = 7.3^2 - 5.5^2\); or \(W/\sin(180° - 41.1°) = 7.3/\sin90°\); or \(W\sin41.1° = 7.3 - 5.5\cos41.1°\) or \(W\cos41.1° = 5.5\sin41.1°\)
A1ft
(if incorrect \(\alpha\))
\(W = 4.8\)
A1
[5]
For using triangle of forces or for resolving in dir° XP₁ or for using Lami's theorem or for resolving forces at X vertically and horizontally (equations must contain not more than one unknown angle) | M1 |
For correct $\Delta$ or resolve XP₁ and $\cos\alpha = \frac{5.5}{7.3}$; or $5.5/\sin(90° + \alpha) = 7.3/\sin90°$ (Lami); or $5.5\cos\alpha + W\sin\alpha = 7.3$ and $5.5\sin\alpha = W\cos\alpha$ | A1 |
Angle AP₁X = 41.1° or 0.718° | A1 |
For correct triangle and $W^2 = 7.3^2 - 5.5^2$; or $W/\sin(180° - 41.1°) = 7.3/\sin90°$; or $W\sin41.1° = 7.3 - 5.5\cos41.1°$ or $W\cos41.1° = 5.5\sin41.1°$ | A1ft | (if incorrect $\alpha$)
$W = 4.8$ | A1 | [5]
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\includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-2_606_843_1731_651}
The diagram shows three particles $A , B$ and $C$ hanging freely in equilibrium, each being attached to the end of a string. The other ends of the three strings are tied together and are at the point $X$. The strings carrying $A$ and $C$ pass over smooth fixed horizontal pegs $P _ { 1 }$ and $P _ { 2 }$ respectively. The weights of $A , B$ and $C$ are $5.5 \mathrm {~N} , 7.3 \mathrm {~N}$ and $W \mathrm {~N}$ respectively, and the angle $P _ { 1 } X P _ { 2 }$ is a right angle. Find the angle $A P _ { 1 } X$ and the value of $W$.
\hfill \mbox{\textit{CAIE M1 2010 Q3 [5]}}