A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal. The box is held in equilibrium by a horizontal force of magnitude 20 N, as shown in Figure 2. The force acts in a vertical plane containing a line of greatest slope of the inclined plane. The box is in equilibrium and on the point of moving down the plane. The box is modelled as a particle. Find (a) the magnitude of the normal reaction of the plane on the box, (b) the coefficient of friction between the box and the plane.

A-Level Edexcel Maths Mechanics Forces: A box of mass 5 kg lies on a rou

A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

Question 3

A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal. The box is held in equilibrium by a horizontal force of magnitude 20 N, as shown in Figur... show full transcript

Worked Solution & Example Answer:A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

Step 1

(a) the magnitude of the normal reaction of the plane on the box

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Answer

To find the normal reaction force, we need to resolve the forces acting on the box perpendicular to the inclined plane.

Identify the forces acting on the box:
- The weight of the box, which acts downwards: $W = mg = 5 imes 9.81 ext{ N} = 49.05 ext{ N}$ .
- The horizontal force of 20 N.
- The normal reaction force, $R$ , acting perpendicular to the inclined surface.
The vertical components of the forces must be balanced since the box is in equilibrium. We can resolve the weight and the horizontal force:
- The component of the weight perpendicular to the inclined plane: $W_ ext{perpendicular} = W imes rac{1}{2} = 49.05 imes rac{ ext{cos}(30^ ext{°})}{ ext{cos}(60^ ext{°})}$ , where $ext{cos}(60^ ext{°}) = 0.5$ .
- The applied force also has a component that affects the normal reaction.
The equation for vertical equilibrium can be set up as: $R = 20 imes ext{cos}(60^ ext{°}) + 49.05 imes ext{cos}(30^ ext{°})$
Solving this gives: $R = 20 imes 0.5 + 49.05 imes rac{ ext{ extsqrt{3}}}{2}$ $R = 10 + 42.43 = 52.4 ext{ N}$

Thus, the magnitude of the normal reaction is approximately 52.4 N.

Step 2

(b) the coefficient of friction between the box and the plane

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Answer

To find the coefficient of friction, we first need to establish the frictional force acting on the box.

The frictional force can be expressed as: $F_f = ext{μ}R$ where $ext{μ}$ is the coefficient of friction.
In addition to the forces, we also need to resolve the forces parallel to the inclined plane. The equation will be: $F_f + 20N ext{ cos}(30^ ext{°}) = 5g ext{ sin}(30^ ext{°})$
The force acting down the inclined plane can be computed as:
- Weight component down the incline: $W_ ext{parallel} = mg ext{ sin}(30^ ext{°})$ $W_ ext{parallel} = 49.05 imes 0.5 = 24.525 ext{ N}$
Now, substituting the values into the equilibrium equation, we get: $F_f + 20 imes 0.866 = 24.525$
Rearranging gives: $F_f = 24.525 - 17.32 ≈ 7.205 ext{ N}$
Now substituting back into the friction equation: $7.205 = ext{μ} imes 52.4$
Solving for $ext{μ}$ , we get: $ext{μ} = rac{7.205}{52.4} \approx 0.137$

Therefore, the coefficient of friction between the box and the plane is approximately 0.137.

A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

Worked Solution & Example Answer:A box of mass 5 kg lies on a rough plane inclined at 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

(a) the magnitude of the normal reaction of the plane on the box

(b) the coefficient of friction between the box and the plane

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