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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

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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\mu R_1, acting leftward, where R1R_1 is the normal force due to the table, which equals 6g.

For the 10 kg mass, the forces are:

  • Tension, T, acting upwards.
  • Weight, W = 10g, acting downwards.
  • Frictional force, F_f = μR2\mu R_2, acting leftward, where R2R_2 equals 10g.

Step 2

Find the common acceleration of the particles.

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Answer

Applying Newton’s second law to both masses:

For the 6 kg mass: 6gT25(6g)=6a6g - T - \frac{2}{5}(6g) = 6a Rearranging gives: T=6g25(6g)6aT = 6g - \frac{2}{5}(6g) - 6a

For the 10 kg mass: T10g+25(10g)=10aT - 10g + \frac{2}{5}(10g) = 10a Substituting from the first equation, we can find: T10g+4g=10aT - 10g + 4g = 10a Solving these equations gives the common acceleration as: a=54m/s2a = \frac{5}{4} \, m/s^2

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: T4g=10a    T=4g+10aT - 4g = 10a \implies T = 4g + 10a Plugging in values gives: T=40+12.5=52.5NT = 40 + 12.5 = 52.5 \, N

Step 4

Comment on the motion of the system if $\mu \geq \frac{2}{3}$.

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Answer

If μ23\mu \geq \frac{2}{3}, 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.T = 30 + 12.5 = 33.75 \, N. Therefore, the tension in the string is 33.75 N.

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