A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal. The package is held in equilibrium by a force of magnitude 45 N acting at an angle of 50° to the plane, as shown in Figure 3. The force is acting in a vertical plane through a line of greatest slope of the plane. The package is modelled as a particle. Find a) the magnitude of the normal reaction of the plane on the package, b) the coefficient of friction between the plane and the package.

A-Level Edexcel Maths Mechanics Forces: A package of mass 4 kg lies on a

A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1

Question 7

A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal. The package is held in equilibrium by a force of magnitude 45 N acting at an angle of... show full transcript

Worked Solution & Example Answer:A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1

Step 1

a) the magnitude of the normal reaction of the plane on the package

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Answer

To find the normal reaction force $R$ , we analyze the forces acting on the package along the plane. The forces include the component of the gravitational force parallel to the plane and the applied force.

The weight of the package is given by: $W = mg = 4 ext{ kg} imes 9.81 ext{ m/s}^2 = 39.24 ext{ N}$
The component of the weight acting perpendicular to the plane is:
$= 39.24 imes 0.866 \ \ = 34.04 ext{ N} $$$
The vertical component of the applied force is:
$= 45 imes 0.766 \ \ = 34.47 ext{ N} $$$
The normal reaction force $R$ can be calculated as:
$= 45 imes 0.766 + 39.24 imes 0.866 \ \ = 34.47 + 34.04 \ \ = 68.51 ext{ N} \ \ \text{Therefore, } R \approx 68.4 ext{ N.}$$$

Step 2

b) the coefficient of friction between the plane and the package

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Answer

To find the coefficient of friction $\mu$ , we use the equilibrium condition along the plane:

The equilibrium equation states: $F = \mu R$ where $F$ is the applied force and $R$ is the normal reaction.
The net force along the plane can be determined by: $F + 4g \text{sin}(30°) = 45 ext{ N} \times ext{cos}(50°)$ Simplifying: $45 ext{ N} \times ext{cos}(50°) - 4g \text{sin}(30°) = 0$ Substituting known values: $45 \times 0.643 - 4 \times 9.81 \times 0.5 = 0 \text{ N}$ $28.935 - 19.62 = 0 \text{ N} \text{ gives us } \mu \approx 0.14.$

Thus, the coefficient of friction is approximately $\mu \approx 0.136$ .

A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1

Worked Solution & Example Answer:A package of mass 4 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 7 - 2008 - Paper 1

a) the magnitude of the normal reaction of the plane on the package

b) the coefficient of friction between the plane and the package

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