OCR MEI Further Mechanics A AS 2019 June — Question 3 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeParticle on inclined plane - force parallel to slope
DifficultyModerate -0.3 This is a standard two-part friction problem on an inclined plane requiring resolution of forces and application of F=μR. Part (a) uses limiting friction with the box about to slip up; part (b) checks if the box slips down when the force is removed. The calculations are straightforward with no conceptual surprises, making it slightly easier than average for A-level mechanics.
Spec3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

3 A box weighing 130 N is on a rough plane inclined at \(12 ^ { \circ }\) to the horizontal.
The box is held at rest on the plane by the action of a force of magnitude 70 N acting up the plane in a direction parallel to a line of greatest slope of the plane.
The box is on the point of slipping up the plane.
  1. Find the coefficient of friction between the box and the plane. The force of magnitude 70 N is removed.
  2. Determine whether or not the box remains in equilibrium.

Question 3:
AnswerMarks Guidance
3(a) WsinF 70
F 70130sin  12 
R  130cos  12  or R130 1  sin  12 2
42.971...

127.159...
AnswerMarks
𝜇 = 0.34M1*
A1
B1
*M1
AnswerMarks
A13.3
1.1
1.1
3.4
AnswerMarks
1.1Resolving forces parallel to the
plane; 3 terms and resolving
essential
soi
F
Using 
AnswerMarks
RCondone sign errors,
wrong use of g and
cos/sin confusion
42.97148019…
127.1591881…
0.3379345…
[5]
AnswerMarks
(b)The weight component (27.0285…N)
is less than the maximum possible frictional
force (42.97…N).
AnswerMarks
Box does remain in equilibriumM1
A13.1b
3.2aAttempt to compare the weight
component parallel to the plane
and the max frictional force. At
least one correct value seen.
AnswerMarks
www130sin12°
May see 15.9(4) > 0
[2]
Question 3:
3 | (a) | WsinF 70
F 70130sin  12 
R  130cos  12  or R130 1  sin  12 2
42.971...

127.159...
𝜇 = 0.34 | M1*
A1
B1
*M1
A1 | 3.3
1.1
1.1
3.4
1.1 | Resolving forces parallel to the
plane; 3 terms and resolving
essential
soi
F
Using 
R | Condone sign errors,
wrong use of g and
cos/sin confusion
42.97148019…
127.1591881…
0.3379345…
[5]
(b) | The weight component (27.0285…N)
is less than the maximum possible frictional
force (42.97…N).
Box does remain in equilibrium | M1
A1 | 3.1b
3.2a | Attempt to compare the weight
component parallel to the plane
and the max frictional force. At
least one correct value seen.
www | 130sin12°
May see 15.9(4) > 0
[2]
3 A box weighing 130 N is on a rough plane inclined at $12 ^ { \circ }$ to the horizontal.\\
The box is held at rest on the plane by the action of a force of magnitude 70 N acting up the plane in a direction parallel to a line of greatest slope of the plane.\\
The box is on the point of slipping up the plane.
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of friction between the box and the plane.

The force of magnitude 70 N is removed.
\item Determine whether or not the box remains in equilibrium.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2019 Q3 [7]}}