Challenging +1.8 This is a challenging statics problem requiring geometric analysis to find contact angles, setting up three equilibrium equations (horizontal/vertical forces and moments), and applying the limiting friction condition. The geometry with the disc and specific distances adds complexity beyond standard mechanics questions, and solving the system requires careful algebraic manipulation across multiple equations, but the underlying principles are standard A-level Further Maths mechanics.
\includegraphics{figure_2}
A uniform smooth disc with centre \(O\) and radius \(a\) is fixed at the point \(D\) on a horizontal surface. A uniform rod of length \(3a\) and weight \(W\) rests on the disc with its end \(A\) in contact with a rough vertical wall. The rod and the disc lie in a vertical plane that is perpendicular to the wall. The wall meets the horizontal surface at the point \(E\) such that \(AE = a\) and \(ED = \frac{5}{4}a\). A particle of weight \(kW\) is hung from the rod at \(B\) (see diagram). The coefficient of friction between the rod and the wall is \(\frac{1}{8}\) and the system is in limiting equilibrium. Find the value of \(k\).
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P: R × AP sin θ + F × AP cos θ – kW × (3a – AP) cos θ
A A
= W × (3a/2 – AP) cos θ)
[ R × 3a/5 + F × 9a/20 – kW × 27a/20 = W × 9a/20 so 12 R +
A A A
9 F – 27 kW = 9 W ]
A
B: R × (3a – AP) – R × 3a sin θ – F × 3a cos θ
P A A
= W × (3a/2) cos θ
[ R × 9a/4 – R × 12a/5 – F × 9a/5 = W × 9a/10
P A A
so 45 R – 48 R – 36 F = 18 W ]
P A A
C: R × (3a/2 – AP) – R × (3a/2) sin θ – F × (3a/2) cos θ + kW
P A A
× (3a/2) cos θ = 0
[ R × 3a/4 – R × 6a/5 – F × 9a/10 + kW × 9a/10 = 0
P A A
so 15 R – 24 R – 18 F + 18 kW = 0 ]
P A A
F: R cos θ × (3a – AP) cos θ – R sin θ × AP sin θ
P P
– F × 3a cos θ = W × (3a/2) cos θ
A
[ (3/5)R × 27a/20 – (4/5)R × 3a/5
P P
– F × 9a/5 = W × 9a/10
A
so 81 R – 48 R – 180 F = 90 W ]
Answer
Marks
Guidance
P P A
M1 A1
F here denotes friction on rod measured in downward
A
dirn;
P denotes point of contact of rod and disc;
θ denotes angle between rod and horizontal.
[AP = 3a/4, sin θ = 4/5, cos θ = 3/5, tan θ = 4/3,
3a – AP = 9a/4, 3a/2 – AP = 3a/4]
See note below on solving question without
introducing R
P
(C denotes mid-point of AB)
(F is vertically below B, on AO extended)
Answer
Marks
Guidance
Question
Answer
Marks
Find two more indep. eqns, e.g. resolution of forces on rod:
Horizontally: R = R sin θ [= 4R /5]
A P P
Vertically: F + (k + 1)W = R cos θ [= 3R /5]
A P P
Along AB: R cos θ = F sin θ + (k + 1)W sin θ
A A
Normal to AB: R = R sin θ + F cos θ + (k + 1)W cos θ
Answer
Marks
Guidance
P A A
B1 B1
A second moment eqn. may be used instead of a
resolution
Count as 2 eqns if used with moments about P (so R
P
absent)
F = +R / 8 or – R / 8 as appropriate
Answer
Marks
Guidance
A A A
B1
Relate F and R (may be implied; and must be
A A
consistent with friction taken down or up in above
eqns)
Answer
Marks
Guidance
[sin θ = 4/5, cos θ = 3/5, tan θ = 4/3]
M1
Eliminate θ from all reqd. independent eqns. for forces
Find either value of k from reqd. independent eqns. for
forces
F ↓: [R = 6W, F = 3W/5, R = 24W/5], k = 2
A P A A
F ↑: [R = 30W/17, F = 3W/17, R = 24W/17], k = 4/17
Answer
Marks
Guidance
A P A A
M1 A1
(or 0⋅235)
Total:
8
Question
Answer
Marks
Question 2:
2 | Take moments for rod about some point such as:
A: R × AP – kW × 3a cos θ = W × (3a/2) cos θ
P
[ R × 3a / 4 – kW × 9a / 5 = W × 9a/10
P
so 15 R – 36 kW = 18 W ]
P
O: F × (5a/4) – kW × (3a cos θ – 5a/4)
A
= – W × (5a/4 – (3a/2) cos θ)
[ F × 5a / 4 – kW × 11a / 20 = – W × 7a/20
A
so 25 F – 11 kW = – 7 W ]
A
P: R × AP sin θ + F × AP cos θ – kW × (3a – AP) cos θ
A A
= W × (3a/2 – AP) cos θ)
[ R × 3a/5 + F × 9a/20 – kW × 27a/20 = W × 9a/20 so 12 R +
A A A
9 F – 27 kW = 9 W ]
A
B: R × (3a – AP) – R × 3a sin θ – F × 3a cos θ
P A A
= W × (3a/2) cos θ
[ R × 9a/4 – R × 12a/5 – F × 9a/5 = W × 9a/10
P A A
so 45 R – 48 R – 36 F = 18 W ]
P A A
C: R × (3a/2 – AP) – R × (3a/2) sin θ – F × (3a/2) cos θ + kW
P A A
× (3a/2) cos θ = 0
[ R × 3a/4 – R × 6a/5 – F × 9a/10 + kW × 9a/10 = 0
P A A
so 15 R – 24 R – 18 F + 18 kW = 0 ]
P A A
F: R cos θ × (3a – AP) cos θ – R sin θ × AP sin θ
P P
– F × 3a cos θ = W × (3a/2) cos θ
A
[ (3/5)R × 27a/20 – (4/5)R × 3a/5
P P
– F × 9a/5 = W × 9a/10
A
so 81 R – 48 R – 180 F = 90 W ]
P P A | M1 A1 | F here denotes friction on rod measured in downward
A
dirn;
P denotes point of contact of rod and disc;
θ denotes angle between rod and horizontal.
[AP = 3a/4, sin θ = 4/5, cos θ = 3/5, tan θ = 4/3,
3a – AP = 9a/4, 3a/2 – AP = 3a/4]
See note below on solving question without
introducing R
P
(C denotes mid-point of AB)
(F is vertically below B, on AO extended)
Question | Answer | Marks | Guidance
Find two more indep. eqns, e.g. resolution of forces on rod:
Horizontally: R = R sin θ [= 4R /5]
A P P
Vertically: F + (k + 1)W = R cos θ [= 3R /5]
A P P
Along AB: R cos θ = F sin θ + (k + 1)W sin θ
A A
Normal to AB: R = R sin θ + F cos θ + (k + 1)W cos θ
P A A | B1 B1 | A second moment eqn. may be used instead of a
resolution
Count as 2 eqns if used with moments about P (so R
P
absent)
F = +R / 8 or – R / 8 as appropriate
A A A | B1 | Relate F and R (may be implied; and must be
A A
consistent with friction taken down or up in above
eqns)
[sin θ = 4/5, cos θ = 3/5, tan θ = 4/3] | M1 | Eliminate θ from all reqd. independent eqns. for forces
Find either value of k from reqd. independent eqns. for
forces
F ↓: [R = 6W, F = 3W/5, R = 24W/5], k = 2
A P A A
F ↑: [R = 30W/17, F = 3W/17, R = 24W/17], k = 4/17
A P A A | M1 A1 | (or 0⋅235)
Total: | 8
Question | Answer | Marks | Guidance
\includegraphics{figure_2}
A uniform smooth disc with centre $O$ and radius $a$ is fixed at the point $D$ on a horizontal surface. A uniform rod of length $3a$ and weight $W$ rests on the disc with its end $A$ in contact with a rough vertical wall. The rod and the disc lie in a vertical plane that is perpendicular to the wall. The wall meets the horizontal surface at the point $E$ such that $AE = a$ and $ED = \frac{5}{4}a$. A particle of weight $kW$ is hung from the rod at $B$ (see diagram). The coefficient of friction between the rod and the wall is $\frac{1}{8}$ and the system is in limiting equilibrium. Find the value of $k$.
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\hfill \mbox{\textit{CAIE FP2 2017 Q2 [8]}}