Mechanics - 5

Classical Mechanics Level 3

Find the sum of the magnitudes of the acceleration of the blocks a m/s 2 a \text{ m/s}^2 , tension in the string T N T \text{ N} , and frictional force acting on B B by the inclined surface f B N f_B \text{ N} . Give your answer to the nearest integer.

Assumptions and Details:

  • No friction between the inclined surface and block A A
  • The coefficient of friction between the inclined surface and block B B is μ B = 0.2 \mu_B = 0.2
  • The string is massless and inextensible.
  • The pulley is massless and there is no friction in the pulley and with the string.
  • Mass of block A A , m A = 3 kg m_A = \text{3 kg}
  • Mass of block B B , m B = 5 kg m_B = \text{5 kg}
  • Angle of incline, θ = sin ⁡ − 1 3 5 \theta = \sin^{-1} \dfrac35
  • Acceleration due to gravity, g = 10 m/s 2 g = \text{10 m/s}^2

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The answer is 28.

Ram Mohith by Ram Mohith , 18, India

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1 solution

Chew-Seong Cheong
Jan 26, 2019

By Newton's second law , we have:

  • The frictional on block B B , f B = μ B m B g cos ⁡ θ ≈ 7.986 N f_B = \mu_B m_B g \cos \theta \approx \text{7.986 N} .
  • The acceleration of the blocks m B g sin ⁡ θ − m A g sin ⁡ θ − f B = ( m A + m B ) a m_Bg \sin \theta - m_Ag \sin \theta - f_B = (m_A + m_B) a ⟹ a = m B g sin ⁡ θ − m A g sin ⁡ θ − f B m A + m B \implies a = \dfrac {m_Bg \sin \theta - m_Ag \sin \theta - f_B}{m_A+m_B} ≈ 0.506 m/s 2 \approx \text{0.506 m/s}^2 .
  • The tension in the string T = m A g sin ⁡ θ + m A a ≈ 19.573 N T = m_Ag\sin \theta + m_Aa \approx \text{19.573 N} .

Therefore, ∣ a ∣ + ∣ T ∣ + ∣ f B ∣ ≈ 28 |a|+|T| + |f_B| \approx \boxed{28} .

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