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A box of mass 1.5 kg is placed on a plane which is inclined at an angle of 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2003 - Paper 1
Question 5
A box of mass 1.5 kg is placed on a plane which is inclined at an angle of 30° to the horizontal. The coefficient of friction between the box and plane is \( \frac{1... show full transcript
Worked Solution & Example Answer:A box of mass 1.5 kg is placed on a plane which is inclined at an angle of 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2003 - Paper 1
Step 1
Step 1: Resolve Forces in the Direction of the Plane
Answer
We can begin by resolving the forces acting on the box. The gravitational force acting on the box can be represented as ( F = m g ), where ( m = 1.5 ) kg and ( g ) is the acceleration due to gravity (approximately ( 9.81 ) m/s²). Therefore, the weight is ( F = 1.5 \times 9.81 = 14.715 ) N.
Next, we resolve this force into components parallel and perpendicular to the inclined plane:
- The component parallel to the plane: ( F_{parallel} = 1.5 g \sin(30°) = 1.5 \times 9.81 \times \frac{1}{2} = 7.36 ) N
- The component perpendicular to the plane: ( F_{normal} = 1.5 g \cos(30°) = 1.5 \times 9.81 \times \frac{\sqrt{3}}{2} = 12.74 ) N
Step 2
Step 2: Apply the Equilibrium Condition in the Normal Direction
Answer
The box is in equilibrium, so the sum of forces in the normal direction must equal zero: [ R - F_{normal} - f = 0 ] Where ( f ) is the frictional force given by ( f = \mu R ), with ( \mu = \frac{1}{5} ). Substituting the values, we get: [ R = 12.74 + \frac{1}{5} R ] This equation can be rearranged to solve for R: [ R - \frac{1}{5} R = 12.74 \Rightarrow \frac{4}{5} R = 12.74 \Rightarrow R = 15.925 \text{ N} ]
Step 3
Step 3: Apply the Equilibrium Condition in the Direction Parallel to the Plane
Answer
Now we apply the equilibrium condition along the inclined plane. The forces acting on the box can be expressed as: [ T \cos(20°) - F_{parallel} - f = 0 ] Substituting the expressions for ( F_{parallel} ) and ( f ), we rewrite the equation as follows: [ T \cos(20°) - 7.36 - \frac{1}{5}R = 0 ] Substituting for R from the previous part, we find: [ T \cos(20°) = 7.36 + \frac{1}{5} \times 15.925 \ \Rightarrow T \cos(20°) = 7.36 + 3.185 \ \Rightarrow T \cos(20°) = 10.545 ]
Step 4