Q3.9 Frictional forces (H)

A mass m1 rests on a smooth inclined plane, of angle θ=30. It is attached (by a rope and pulley) to a mass m2=10 kg which rests on a horizontal surface with which its coefficients of friction are μs=0.5 and μk=0.25.
Graphic - No title - twomassfriction
  1. Find the maximum value of m1 possible without the system sliding.
  2. Find the acceleration of the system if it is prompted to slide, when m1 has the value found in the preceding part of the question.
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Hint

In each case you will need to write down Newton’s 2nd law applied, separately, to each of the masses.
 

Solution

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Solution

  1. Let the tension in the rope be T. Since the system is in static equilibrium, the net force on each mass must vanish.

    Mass m1 experiences two forces parallel to the slope:

    1. an up-the-slope force of magnitude T exerted by the rope
    2. a down-the-slope force of magnitude m1gsinθ due to its weight

    Note that (life is kind, sometimes) the slope itself is supposed smooth so that no frictional force acts on m1. Newton’s 2nd law implies then

    T=m1gsinθ

    Mass m2 experiences two forces in the horizontal direction:

    1. a leftward force of magnitude T exerted by the rope
    2. a rightward force due to friction of magnitude
      FF=μsFN=μsm2g

    Note that in setting FF=μsFN we are picking out the circumstances in which the system is on the point of slipping.

    Newton’s 2nd law applied to this mass then implies

    T=μsm2g
    Combining these two results (to eliminate T) we get
    m1gsinθ=μsm2g
    implying
    \[ m_1= \frac{\mu_s m_2}{\sin\theta} = 10 \,kg \]

  2. Suppose now that the system is nudged so it begins to slide. Denote the tension now by T (to emphasize it is actually different; but we are going to eliminate it anyway). Let a denote the acceleration of the mass m1 down the plane, and also of the mass m2 to the left. (Since these masses are connected by a rope their accelerations must be related in this way).

    Applying Newton’s 2nd law to mass m1 gives

    m1gsinθ-T=m1a
    Applying it to mass m2 gives
    \[ m_2a =T^{\prime} -F^{\prime}_F = T^{\prime}-\mu_k m_2 g \]
    where, now, the coefficient of kinetic friction needs to be used to characterise the size of the frictional force. Combining these equations to eliminate T we find
    m1(gsinθ-a)=T=m2(a+μkg)
    Reorganising gives
    a(m1+m2)=g(m1sinθ-m2μk)
    Thus
    \[ a= \frac{g(m_1 \sin \theta -m_2 \mu_k)}{m_1+m_2} = \frac{g}{8} \]