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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
Find the magnitude of the normal reaction of the plane on the box
Answer
To find the normal reaction ( R ) acting on the box, we resolve the forces perpendicular to the inclined plane.
The forces in the vertical direction include the weight of the box and the vertical component of the applied force:
- Weight of the box: ( W = mg = 5 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 49.05 , \text{N} )
- The vertical component of the applied force: ( F_v = 20 , \text{N} \cdot \cos(30°) = 20 \times \frac{\sqrt{3}}{2} \approx 17.32 , \text{N} )
Now we express the forces in equilibrium:
[ R = W \cdot \cos(30°) + F_v \cdot \sin(30°) ] Substituting values: [ R = 49.05 \cdot \cos(30°) + 17.32 \cdot \sin(30°) ] Evaluating: [ R = 49.05 \times \frac{\sqrt{3}}{2} + 17.32 \times \frac{1}{2} \approx 52.4 , \text{N} ]
Thus, the magnitude of the normal reaction is approximately 52.4 N.
Step 2
Find the coefficient of friction between the box and the plane
Answer
To find the coefficient of friction ( \mu ), we first consider the forces acting parallel to the inclined plane.
Using the formula for friction: [ F_f = \mu R ]
The forces acting down the incline include the component of the weight of the box down the slope and the horizontal force's component:
- The component of the weight down the incline: ( W_\text{down} = W \cdot \sin(30°) = 49.05 \cdot \frac{1}{2} = 24.525 , \text{N} )
- The horizontal force's component: ( F_h = 20 , \text{N} \cdot \sin(30°) = 20 \times \frac{1}{2} = 10 , \text{N} )
Thus, the total force down the incline is: [ F_\text{total} = W_\text{down} - F_h = 24.525 - 10 = 14.525 , \text{N} ]
Setting the frictional force equal to the total force: [ 14.525 = \mu \cdot 52.4 ] Solving for ( \mu ): [ \mu = \frac{14.525}{52.4} \approx 0.138 ]
Therefore, the coefficient of friction between the box and the plane is approximately 0.138.