Challenging +1.8 This is a challenging statics problem requiring resolution of forces in two directions, taking moments about a strategic point, using the friction law at limiting equilibrium, and working with the geometry of the string constraint. It involves multiple simultaneous equations and careful geometric reasoning with the given tan θ = 2 condition. While the techniques are standard A-level mechanics (moments, resolution, friction), the multi-step coordination and geometric setup place it well above average difficulty for A-level, though not at the extreme end since the approach is systematic once the setup is understood.
\includegraphics{figure_4}
A uniform rod \(AB\) of length \(4a\) and weight \(W\) rests with the end \(A\) in contact with a rough vertical wall. A light inextensible string of length \(\frac{5}{2}a\) has one end attached to the point \(C\) on the rod, where \(AC = \frac{3}{2}a\). The other end of the string is attached to a point \(D\) on the wall, vertically above \(A\). The vertical plane containing the rod \(AB\) is perpendicular to the wall. The angle between the rod and the wall is \(\theta\), where \(\tan \theta = 2\) (see diagram). The end \(A\) of the rod is on the point of slipping down the wall and the coefficient of friction between the rod and the wall is \(\mu\).
Find, in either order, the tension in the string and the value of \(\mu\). [10]
C: F × 5a/2 sin θ – R × 5a/2 cos θ – W × ½ a sin θ = 0
A A
B: F × 4a sin θ – R × 4a cos θ – W × 2a sin θ
A A
+ T × 3a/2 sin (π – 2θ) = 0
G: F × 2a sin θ – R × 2a cos θ – T × ½ a sin (π – 2θ) = 0
A A
(G is mid-point of AB)
D: R × 5a cos θ – W × 2a sin θ = 0
Answer
Marks
Guidance
A
M1 A1
Take moments for rod about one chosen point
[sin θ = 2/√5, cos θ = 1/√5, sin (π – 2θ) = sin 2θ = 4/5]
R = T sin θ
A
F = W – T cos θ
Answer
Marks
Guidance
A
B1
Find two more independent equations
B1
e.g. resolution of forces on rod (a second moment equation may be
used)
F = µR
Answer
Marks
Guidance
A A
B1
Relate F and R (may be implied)
A A
Answer
Marks
Guidance
T = (2W sin θ ) / (5/2 sin 2θ ) = 2W / (5 cos θ )
M1
Find T by any method (e.g. from moments about A)
= 2W/√5 or (2√5/5) W or 0⋅894 W
A1
F = (3/5) W and R = (4/5) W
A A
Answer
Marks
Guidance
µ = ¾ or 0⋅75
M1 A1
Find or imply F and R by any method (e.g. from resolutions)
A A
and hence µ
A1
10
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | A: T × 5a/2 sin (π – 2θ) – W × 2a sin θ = 0
C: F × 5a/2 sin θ – R × 5a/2 cos θ – W × ½ a sin θ = 0
A A
B: F × 4a sin θ – R × 4a cos θ – W × 2a sin θ
A A
+ T × 3a/2 sin (π – 2θ) = 0
G: F × 2a sin θ – R × 2a cos θ – T × ½ a sin (π – 2θ) = 0
A A
(G is mid-point of AB)
D: R × 5a cos θ – W × 2a sin θ = 0
A | M1 A1 | Take moments for rod about one chosen point
[sin θ = 2/√5, cos θ = 1/√5, sin (π – 2θ) = sin 2θ = 4/5]
R = T sin θ
A
F = W – T cos θ
A | B1 | Find two more independent equations
B1 | e.g. resolution of forces on rod (a second moment equation may be
used)
F = µR
A A | B1 | Relate F and R (may be implied)
A A
T = (2W sin θ ) / (5/2 sin 2θ ) = 2W / (5 cos θ ) | M1 | Find T by any method (e.g. from moments about A)
= 2W/√5 or (2√5/5) W or 0⋅894 W | A1
F = (3/5) W and R = (4/5) W
A A
µ = ¾ or 0⋅75 | M1 A1 | Find or imply F and R by any method (e.g. from resolutions)
A A
and hence µ
A1
10
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
A uniform rod $AB$ of length $4a$ and weight $W$ rests with the end $A$ in contact with a rough vertical wall. A light inextensible string of length $\frac{5}{2}a$ has one end attached to the point $C$ on the rod, where $AC = \frac{3}{2}a$. The other end of the string is attached to a point $D$ on the wall, vertically above $A$. The vertical plane containing the rod $AB$ is perpendicular to the wall. The angle between the rod and the wall is $\theta$, where $\tan \theta = 2$ (see diagram). The end $A$ of the rod is on the point of slipping down the wall and the coefficient of friction between the rod and the wall is $\mu$.
Find, in either order, the tension in the string and the value of $\mu$. [10]
\hfill \mbox{\textit{CAIE FP2 2019 Q4 [10]}}