Challenging +1.8 This is a challenging statics problem requiring geometric analysis to find angles and contact points, setting up moment equations about multiple points, resolving forces in two directions, applying limiting friction conditions, and solving a system of equations. The geometry is non-trivial (finding where the rod touches the disc) and the problem requires careful bookkeeping of multiple forces and distances, but follows standard mechanics methodology without requiring novel insight.
\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 \(ED = \frac{3}{4}a\). 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 \(k\).
[8]
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 $ED = \frac{3}{4}a$. 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 $k$.
[8]
\hfill \mbox{\textit{CAIE FP2 2017 Q2 [8]}}