Edexcel M1 — Question 7 14 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeString at angle to slope
DifficultyStandard +0.3 This is a standard M1 equilibrium problem on an inclined plane with a string at an angle. Part (a) is routine resolution of forces parallel to the plane. Part (b) requires perpendicular resolution. Parts (c) and (d) extend to friction but remain straightforward applications of equilibrium principles. The multi-part structure and 3D force resolution make it slightly above average difficulty, but all techniques are standard M1 content with no novel problem-solving required.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

\includegraphics{figure_3} Figure 3 shows a block of mass 25 kg held in equilibrium on a plane inclined at an angle of 35° to the horizontal by means of a string which is at an angle of 15° to the line of greatest slope of the plane. In an initial model of the situation, the plane is assumed to be smooth. Giving your answers correct to 3 significant figures,
  1. show that the tension in the string is 145 N. [3 marks]
  2. find the magnitude of the reaction between the plane and the block. [4 marks]
In a more refined model, the plane is assumed to be rough. Given that the tension in the string can be increased to 200 N before the block begins to move up the slope,
  1. find, correct to 3 significant figures, the magnitude of the frictional force and state the direction in which it acts. [4 marks]
  2. Without performing any further calculations, state whether the reaction calculated in part (b) will increase, decrease or remain the same in the refined model. Give a reason for your answer. [3 marks]

(a) resolve // to plane: \(T\cos 15° - 25\sin 35° = 0\)
AnswerMarks Guidance
so \(T = 145.48 = 145\) N (3sf)M2 A1
(b) resolve perp. to plane: \(R + T\sin 15° - 25\cos 35° = 0\)
AnswerMarks Guidance
\(R = 25\cos 35° - 145.48 \times \sin 15° = 163.038 = 163\) N (3sf)M1 A1 M1 A1
(c) resolve // to plane: \(200\cos 15° - F - 25\sin 35° = 0\)
AnswerMarks Guidance
\(F = 52.7\) N (3sf) down the slopeM2 A2
(d) decrease
AnswerMarks Guidance
e.g. perp. to plane, same force "down", more from T "up" ∴ R lessB1 B2
Total: (75)
**(a)** resolve // to plane: $T\cos 15° - 25\sin 35° = 0$
so $T = 145.48 = 145$ N (3sf) | M2 | A1 |

**(b)** resolve perp. to plane: $R + T\sin 15° - 25\cos 35° = 0$
$R = 25\cos 35° - 145.48 \times \sin 15° = 163.038 = 163$ N (3sf) | M1 A1 | M1 A1 |

**(c)** resolve // to plane: $200\cos 15° - F - 25\sin 35° = 0$
$F = 52.7$ N (3sf) down the slope | M2 | A2 |

**(d)** decrease
e.g. perp. to plane, same force "down", more from T "up" ∴ R less | B1 | B2 | (14) |

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**Total: (75)**
\includegraphics{figure_3}

Figure 3 shows a block of mass 25 kg held in equilibrium on a plane inclined at an angle of 35° to the horizontal by means of a string which is at an angle of 15° to the line of greatest slope of the plane.

In an initial model of the situation, the plane is assumed to be smooth. Giving your answers correct to 3 significant figures,

\begin{enumerate}[label=(\alph*)]
\item show that the tension in the string is 145 N. [3 marks]
\item find the magnitude of the reaction between the plane and the block. [4 marks]
\end{enumerate}

In a more refined model, the plane is assumed to be rough.

Given that the tension in the string can be increased to 200 N before the block begins to move up the slope,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find, correct to 3 significant figures, the magnitude of the frictional force and state the direction in which it acts. [4 marks]
\item Without performing any further calculations, state whether the reaction calculated in part (b) will increase, decrease or remain the same in the refined model. Give a reason for your answer. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q7 [14]}}