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 Moments: 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 can analyze the forces acting on the package. The weight of the package $W = mg = 4 imes 9.81 = 39.24 ext{ N}$ acts vertically downward. The normal reaction $R$ acts perpendicular to the surface.

We resolve the 45 N force into components parallel and perpendicular to the inclined plane:

The component of the 45 N force acting perpendicular to the plane is: $F_{N} = 45 imes ext{cos}(50°)$
The component of the gravitational force acting perpendicular to the plane is: $W_{N} = 4g imes ext{cos}(30°) = 4 imes 9.81 imes ext{cos}(30°)$

Using equilibrium in the normal direction: $R = 45 imes ext{cos}(50°) + 4g imes ext{cos}(30°)$ Calculating this: $R ext{ approximately equals } 68.4 ext{ N}$ Thus, the magnitude of the normal reaction of the plane on the package is approximately 68 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 for forces along the plane:

$F_{f} + W_{ ext{parallel}} = F_{N}$

Where:

The frictional force $F_{f} = \mu R$
The component of the weight acting parallel to the plane is: $W_{ ext{parallel}} = 4g imes ext{sin}(30°)$

Substituting into the equilibrium equation gives: $\mu R + 4g imes ext{sin}(30°) = 45 imes ext{cos}(50°)$

Rearranging to find $\mu$ : $\mu = \frac{45 imes ext{cos}(50°) - 4g imes ext{sin}(30°)}{R}$

Substituting in values where $R$ is previously calculated: $\mu = \frac{45 imes ext{cos}(50°) - 4g imes ext{sin}(30°)}{68.4}$ Calculating gives: $\mu \text{ approximately equals } 0.136$ Therefore, the coefficient of friction between the plane and the package is approximately 0.14.

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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