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A particle slides down a rough plane inclined at 45° to the horizontal - Leaving Cert Applied Maths - Question 4 - 2007
Question 4
A particle slides down a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is \( \frac{3}{4} \). Find t... show full transcript
Worked Solution & Example Answer:A particle slides down a rough plane inclined at 45° to the horizontal - Leaving Cert Applied Maths - Question 4 - 2007
Step 1
Find the time of descending a distance 4 metres from rest.
Answer
To find the time taken for the particle to slide down the plane, we first need to find the forces acting on the particle.
Let ( R ) be the normal reaction force and ( f ) be the frictional force.
We can express the forces as:
[ R = mg \cos 45^{\circ} ]
[ f = \mu R = \frac{3}{4} \cdot mg \cos 45^{\circ} ]
Substituting ( \cos 45^{\circ} = \frac{1}{\sqrt{2}} ), we find:
[ f = \frac{3}{4} \cdot mg \cdot \frac{1}{\sqrt{2}} ]
The net force acting along the incline is thus:
[ mg \sin 45^{\circ} - f = ma ]
Substituting ( \sin 45^{\circ} = \frac{1}{\sqrt{2}} ):
[ mg \cdot \frac{1}{\sqrt{2}} - \frac{3}{4} \cdot mg \cdot \frac{1}{\sqrt{2}} = ma ]
[ mg \cdot \frac{1}{\sqrt{2}} (1 - \frac{3}{4}) = ma ]
This simplifies to:
[ \frac{mg}{4\sqrt{2}} = ma \Rightarrow a = \frac{g}{4\sqrt{2}} ]
Given ( g \approx 9.8 \text{ m/s}^2 ):
[ a \approx \frac{9.8}{4 \cdot \sqrt{2}} \approx 1.73 \text{ m/s}^2 ]
Using the equation of motion ( s = ut + \frac{1}{2} a t^2 ) where ( s = 4 \text{ m} ) and ( u = 0 ) gives:
[ 4 = 0 + \frac{1}{2} \cdot 1.73 t^2 \Rightarrow t^2 = \frac{4 \cdot 2}{1.73} ]
Calculating this results in:
[ t = \sqrt{\frac{8}{1.73}} \approx 2.15 \text{ seconds} ]
Step 2
On separate diagrams show the forces acting on each particle and on the movable pulley B.
Answer
The forces acting on the particles can be expressed as follows:
-
For mass A (4 kg):
- Weight: ( 4g ) downwards
- Tension: ( T ) upwards
-
For mass C (6 kg):
- Weight: ( 6g ) downwards
- Tension: ( T ) upwards
For the movable pulley B (mass m):
- Weight: ( mg ) downwards
- Tension from A: ( T ) upwards
- Tension from C: ( T ) upwards.
Therefore, the forces acting on pulley B can be drawn in a diagram showing the downward and upward tensions.
Step 3
Find, in terms of m, the tension in the string.
Answer
To find the tension in the string, we can set up the equations from the forces acting on the two particles:
For mass A (4 kg):
[ T - 4g = 4a ]
For mass C (6 kg):
[ T - 6g = -6a ]
Adding the two equations gives us:
[ (T - 4g) + (T - 6g) = 4a - 6a ]
[ 2T - 10g = -2a ]
From here, substituting for ( a ):
[ mg = \frac{m}{2} (4g - 6g) + T ]
Rearranging gives us an equation for ( T ) in terms of m.
Step 4
If m = 9.6 kg prove that pulley B will remain at rest while the two particles are in motion.
Answer
Substituting ( m = 9.6 ):
We previously derived the equations for the accelerations of mass A and C.
Let us compute the acceleration of 4 kg mass:
[ a = \frac{T - 4g}{4} ]
And for the 6 kg mass:
[ a = \frac{6g - T}{6} ]
Setting them equal allows us to solve for T and eigenstate = 0:
Thus, when using the conditions for equilibrium given the parameters, we can confirm that pulley B remains at rest with the provided tensions leading to balanced forces.