A particle of mass 10 kg is connected to another particle of mass 6 kg by a taut light inelastic string which passes over a smooth light pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2017
Question 4
A particle of mass 10 kg is connected to another particle of mass 6 kg by a taut light inelastic string which passes over a smooth light pulley at the edge of a roug... show full transcript
Worked Solution & Example Answer:A particle of mass 10 kg is connected to another particle of mass 6 kg by a taut light inelastic string which passes over a smooth light pulley at the edge of a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2017
Step 1
Show on separate diagrams the forces acting on each particle.
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Answer
For the 6 kg mass, the forces are:
Tension, T, acting upwards.
Weight, W = 6g, acting downwards.
Frictional force, F_f = μR1, acting leftward, where R1 is the normal force due to the table, which equals 6g.
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Answer
Applying Newton’s second law to both masses:
For the 6 kg mass:
6g−T−52(6g)=6a
Rearranging gives:
T=6g−52(6g)−6a
For the 10 kg mass:
T−10g+52(10g)=10a
Substituting from the first equation, we can find:
T−10g+4g=10a
Solving these equations gives the common acceleration as:
a=45m/s2
Step 3
Find the tension in the string.
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Answer
Using the common acceleration:
Substituting into one of the equations to find T:
T−4g=10a⟹T=4g+10a
Plugging in values gives:
T=40+12.5=52.5N
Step 4
Comment on the motion of the system if $\mu \geq \frac{2}{3}$.
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Answer
If μ≥32, then the frictional force would be greater than the gravitational force acting on the 10 kg mass.
This would result in no net motion in the system, hence the system would remain stationary.
Step 5
Find (i) the common acceleration of the masses.
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Answer
Applying Newton's second law to each mass:
For the 5 kg mass:
\sin(\alpha) = \frac{3}{5}.$$
This gives:
$$3g - T = 5a\implies T = 3g - 5a.$$
For the 3 kg mass:
$$5g - T = 3a$$
Thus, manipulating these equations leads to:
The common acceleration calculated is:
$$a = \frac{5}{4} \, m/s^2$$.
Step 6
Find (ii) the tension in the string.
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Answer
Using the values obtained:
From previous equations:
Substituting the common acceleration back into the equation for tension, we find:
T=30+12.5=33.75N.
Therefore, the tension in the string is 33.75 N.
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