Question 4 - A-Level Maths

OCR MEI M1 2010 January — Question 4 8 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Year	2010
Session	January
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Equilibrium of particle under coplanar forces
Difficulty	Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions. Part (i) is simple trigonometry (20cos15°), parts (ii)-(iii) involve standard force balance equations, and part (iv) tests conceptual understanding that horizontal equilibrium is independent of vertical forces. All techniques are routine M1 content with no novel problem-solving required.
Spec	3.03e Resolve forces: two dimensions 3.03m Equilibrium: sum of resolved forces = 0

4 A box of mass 2.5 kg is on a smooth horizontal table, as shown in Fig. 4. A light string AB is attached to the table at A and the box at B . AB is at an angle of \(50 ^ { \circ }\) to the vertical. Another light string is attached to the box at C ; this string is inclined at \(15 ^ { \circ }\) above the horizontal and the tension in it is 20 N . The box is in equilibrium. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{eafaf02f-bcd4-4368-a282-61ef1ad074da-3_403_1063_1085_539} \captionsetup{labelformat=empty} \caption{Fig. 4}

\end{figure}

Calculate the horizontal component of the force exerted on the box by the string at C .
Calculate the tension in the string AB .
Calculate the normal reaction of the table on the box. The string at C is replaced by one inclined at \(15 ^ { \circ }\) below the horizontal with the same tension of 20 N .
Explain why this has no effect on the tension in string AB .

Show mark scheme Show mark scheme source

Question 4:

Part (i):

Answer	Marks	Guidance
Answer	Mark	Guidance
\(20\cos 15 = 19.3185...\) so \(19.3\) N (3 s.f.) in direction BC	B1	Accept no direction. Must be evaluated

Part (ii):

Answer	Marks	Guidance
Answer	Mark	Guidance
Let tension be \(T\); \(T\sin 50 = 19.3185...\)	M1	Accept \(\sin \leftrightarrow \cos\) but not (i) \(\times \sin 50\)
so \(T = 25.2185...\) so \(25.2\) N (3 s.f.)	F1	FT their \(19.3...\) only. cwo

Part (iii):

Answer	Marks	Guidance
Answer	Mark	Guidance
\(R + 20\sin 15 - 2.5g - 25.2185...\times \cos 50 = 0\)	M1	Allow 1 force missing or 1 tension not resolved. FT \(T\). No extra forces. Accept mass used. Accept \(\sin \leftrightarrow \cos\)
B1	Weight correct
A1	All correct except sign errors. FT their \(T\)
\(R = 35.5337...\) so \(35.5\) N (3 s.f.)	A1	cao. Accept 35 or 36 for 2 s.f.

Part (iv):

Answer	Marks	Guidance
Answer	Mark	Guidance
The horizontal resolved part of the 20 N force is not changed.	E1	Accept no reference to vertical component but do not accept 'no change' to both components. No need to be explicit that value of tension in AB depends only on horizontal component of force at C

# Question 4:

## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $20\cos 15 = 19.3185...$ so $19.3$ N (3 s.f.) in direction BC | B1 | Accept no direction. Must be evaluated |

## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Let tension be $T$; $T\sin 50 = 19.3185...$ | M1 | Accept $\sin \leftrightarrow \cos$ but not (i) $\times \sin 50$ |
| so $T = 25.2185...$ so $25.2$ N (3 s.f.) | F1 | FT **their** $19.3...$ only. cwo |

## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R + 20\sin 15 - 2.5g - 25.2185...\times \cos 50 = 0$ | M1 | Allow 1 force missing or 1 tension not resolved. FT $T$. No extra forces. Accept mass used. Accept $\sin \leftrightarrow \cos$ |
| | B1 | Weight correct |
| | A1 | All correct except sign errors. FT **their** $T$ |
| $R = 35.5337...$ so $35.5$ N (3 s.f.) | A1 | cao. Accept 35 or 36 for 2 s.f. |

## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| The horizontal resolved part of the 20 N force is not changed. | E1 | Accept no reference to vertical component but do not accept 'no change' to both components. No need to be explicit that value of tension in AB depends only on horizontal component of force at C |

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Show LaTeX source

4 A box of mass 2.5 kg is on a smooth horizontal table, as shown in Fig. 4. A light string AB is attached to the table at A and the box at B . AB is at an angle of $50 ^ { \circ }$ to the vertical. Another light string is attached to the box at C ; this string is inclined at $15 ^ { \circ }$ above the horizontal and the tension in it is 20 N . The box is in equilibrium.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{eafaf02f-bcd4-4368-a282-61ef1ad074da-3_403_1063_1085_539}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

(i) Calculate the horizontal component of the force exerted on the box by the string at C .\\
(ii) Calculate the tension in the string AB .\\
(iii) Calculate the normal reaction of the table on the box.

The string at C is replaced by one inclined at $15 ^ { \circ }$ below the horizontal with the same tension of 20 N .\\
(iv) Explain why this has no effect on the tension in string AB .

\hfill \mbox{\textit{OCR MEI M1 2010 Q4 [8]}}

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