Question 3:
3 | (a) | Contact force at A resolved into separate normal and frictional
components; allow friction either vertically upwards/downwards.
80 N force vertically downward at centre of beam (G) and
contact force at C perp. to beam in the correct direction. | B1
B1
[2] | 3.3
3.3 | Forces must have arrows and be appropriately labelled
No extra forces, but condone components shown as well
(Forces at A) Accept contact force as a single force
(neither horizontal nor vertical nor parallel to beam)
B0 for friction labelled 𝜇𝑅
A
(Forces elsewhere) Must be clear right angle or with
right angle marker present.
Max B1 if same label used for two forces
(b) | (i) | Take the frictional force at A to be F vertically downwards.
Resolving vertically: R s i n 6 0  = F + 8 0
C
Moments about A: 𝑅 (2) = 80(3.5sin60°)
C
[ 𝑅 = 70√3 = 121.24 ]
C
[ 105 = 𝐹+80 ]
F = 2 5 | M1
M1
A1
[3] | 1.1
1.1
2.1 | Resolving to obtain an equation involving F, e.g.
Parallel to AB: 𝑅 sin60° = 80cos60°+𝐹cos60°
A
Perp to AB: 𝑅 = 80sin60°+𝐹sin60°+𝑅 cos60°
C A
A moments equation, e.g. about
C: 𝑅 (2cos60°)+𝐹(2sin60°) = 80(1.5sin60°)
A
G: 𝑅 (3.5cos60°)+𝐹(3.5sin60°) = 𝑅 (1.5)
A C
B: 𝑅 (7cos60°)+𝐹(7sin60°)+80(3.5sin60°)=𝑅 (5)
A C
No omitted or extra terms. Allow sign errors and sin/cos
confusion. Allow 𝑅 in wrong direction (e.g. vertical).
C
In moments about C, note that 2cos60° = 1
AG Fully correct working
With F upwards leading to 𝐹 = −25 they must say ‘so
the magnitude is 25’ or ‘i.e. 25 downwards’ to earn A1
3
cont | (b) | (ii) | 𝑅 = 𝑅 cos60°
A C
R =35 3( = 6 0 .6 2 1 7 )
A
( ) 2
Magnitude = 2 5 2 + 3 5 3
Magnitude is 65.6 (N) | M1*
M1dep*
A1
[3] | 1.1
1.1
1.1 | Resolving horizontally
OR Obtaining another resolving or moments equation,
involving 𝑅 , not credited in (i)
A
OR Obtaining 𝑅 from equation(s) credited in (i),
A
possibly using 𝐹 = (±)25
This M1 can be awarded for work done in (i)
Obtaining a value for 𝑅 and using F 2 + R 2
A A
Accept 4 3 0 0
(c) | 25
tan𝜃 =
𝑅A
22.4(109…)° below the horizontal. | M1
A1
[2] | 1.1
2.5 | Equation for a relevant angle
𝑅A 25
e.g. tan𝛼 = , sin𝜃 = etc
25 65.6
Provided magnitude in (b)(ii) has been calculated from
perpendicular components, allow M1 for using the same
components to find an angle
Or 67.5(890…)° from the downward vertical
Direction must be clear
(d) | FR  
A
5√3
𝜇 ≥ , 𝜇 ≥ 0.412
21 | M1
A1 FT
[2] | 3.4
1.1 | 𝐹 = 𝜇𝑅 or 𝐹 ≤ 𝜇𝑅 seen or implied
A A
Must be their forces at A (on diagram) or their values
25
FT is 𝜇 ≥
their 𝑅A
Allow exact or approximate form. Allow >
Allow inequality stated in words.
Ignore upper limit, e.g. A1 for 0.412 ≤ 𝜇 ≤ 1
(e) | Both F and R would increase so the magnitude must increase.
A | B1
[1] | 2.4 | Must acknowledge that both components increase.
OR Other reasonable explanation, e.g.
‘Moment of weight (about C) increases, so the total
contact force must increase to balance it’