Challenging +1.2 This is a standard statics problem requiring resolution of forces and taking moments about a point, with smooth contacts simplifying the force directions. The constraint tan α = 2tan θ adds a geometric relationship to work with, but the solution follows a systematic approach: resolve horizontally/vertically, take moments, and use the given constraint. While it requires careful bookkeeping across multiple steps (9 marks), it's a textbook application of equilibrium principles without requiring novel insight or particularly challenging manipulation.
\includegraphics{figure_1}
A uniform ladder \(AB\), of length \(3a\) and weight \(W\), rests with the end \(A\) in contact with smooth horizontal ground and the end \(B\) against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point \(C\), where \(AC = a\). The other end of the rope is fixed to the point \(D\) at the base of the wall and the rope \(DC\) is in the same vertical plane as the ladder \(AB\). The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle \(\theta\) with the horizontal and the rope making an angle \(\alpha\) with the horizontal (see diagram). It is given that \(\tan \alpha = 2\tan \theta\). Find, in terms of \(W\) and \(\alpha\), the tension in the rope and the magnitudes of the forces acting on the ladder at \(A\) and at \(B\). [9]
Take moments about A: RB 3a sin θ = W (3a/2) cos θ
(a may be omitted from moment eqns) + T a(sin α cos θ + cos α sin θ )
or + T a sin (α + θ )
or + T 3a cos θ sin α M1 A1
Take moments about B: RA3a cos θ = W (3a/2) cos θ
+ T 2a(sin α cos θ + cos α sin θ )
or + T 2a sin (α + θ )
or + T 3a sin θ cos α (M1 A1)
Take moments about C: RA a cos θ + W (a/2) cos θ
= RB 2a sin θ (M1 A1)
Take moments about D: RA 3a cos θ – W (3a/2) cos θ
= RB 3a sin θ (M1 A1)
Solve for T, RA, RB (AEF in W and α): T = W / 2 sin α or ½W cosec α B1
RA = 3W / 2 B1
Answer
Marks
Guidance
RB = W / 2 tan α or ½W cot α B1
9
9
Question 1:
1 | Find 3 independent equations for T, RA, RB :
Resolve horizontally: RB = T cos α M1 A1
Resolve vertically: RA = W + T sin α M1 A1
Take moments about A: RB 3a sin θ = W (3a/2) cos θ
(a may be omitted from moment eqns) + T a(sin α cos θ + cos α sin θ )
or + T a sin (α + θ )
or + T 3a cos θ sin α M1 A1
Take moments about B: RA3a cos θ = W (3a/2) cos θ
+ T 2a(sin α cos θ + cos α sin θ )
or + T 2a sin (α + θ )
or + T 3a sin θ cos α (M1 A1)
Take moments about C: RA a cos θ + W (a/2) cos θ
= RB 2a sin θ (M1 A1)
Take moments about D: RA 3a cos θ – W (3a/2) cos θ
= RB 3a sin θ (M1 A1)
Solve for T, RA, RB (AEF in W and α): T = W / 2 sin α or ½W cosec α B1
RA = 3W / 2 B1
RB = W / 2 tan α or ½W cot α B1 | 9 | 9
\includegraphics{figure_1}
A uniform ladder $AB$, of length $3a$ and weight $W$, rests with the end $A$ in contact with smooth horizontal ground and the end $B$ against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point $C$, where $AC = a$. The other end of the rope is fixed to the point $D$ at the base of the wall and the rope $DC$ is in the same vertical plane as the ladder $AB$. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle $\theta$ with the horizontal and the rope making an angle $\alpha$ with the horizontal (see diagram). It is given that $\tan \alpha = 2\tan \theta$. Find, in terms of $W$ and $\alpha$, the tension in the rope and the magnitudes of the forces acting on the ladder at $A$ and at $B$. [9]
\hfill \mbox{\textit{CAIE FP2 2015 Q1 [9]}}