Fixed Pulley

Task number: 1986

A weight with the mass of 0.55 kg and a bucket with the mass of 0.45 kg. hang on a fixed pulley. Determine the acceleration of the system and the force with which the pulley acts upon its axis. Neglect the friction and the mass of the pulley and the string.

Figure to the assignment
  • Given values

    m1 = 0.55 kg mass of the weight
    m2 = 0.45 kg mass of the bucket
    a = ? m·s−2 acceleration of the system
    Fk = ? N force with which the pulley acts upon its axis.
  • Hint 1 – Equations of motion, calculation of acceleration

    Draw a free body diagram and mark all the forces acting upon the weight and the bucket into the diagram. Write the equations of motion for the bucket and the weight. Choose a coordinate system.

  • Hint 2 – Force acting upon the axis of the pulley

    The force with which the pulley acts upon its axis is, according to Newton's third law, equal to the force with which the axis acts upon the pulley. Draw all the forces acting upon the pulley and realise what the net force is.

  • OVERALL SOLUTION

    Forces acting upon the weights:

    Weight m1:

    \(\vec{F}_\mathrm{G1}\)…gravitational force

    \(\vec{T}_1\)…tensile force of the string

    Bucket m2:

    \(\vec{F}_\mathrm{G2}\)…gravitational force

    \(\vec{T}_2\)…tensile force of the string

    Forces acting upon the bodies

    Equations of motion:

    Weight m1:

    \[\vec{F}_\mathrm{G1}+\vec{T}_1\,=\,m_1\vec{a}.\]

    Bucket m2:

    \[\vec{F}_\mathrm{G2}+\vec{T}_2\,=\,m_2\vec{a}.\]

    The weight and the bucket move with the same acceleration in opposite directions.

    Forces acting upon the bodies (with coordinate axes)

    We choose the y axis as depicted in the picture. We rewrite the equations using scalars:

    \[F_\mathrm{G1} - T_1 \,=\, m_1a,\tag{1}\] \[T_2 - F_\mathrm{G2} \,=\, m_2a.\tag{2}\]

    Because we neglect the mass of the pulley, it has no moment of inertia and does not affect the forces of the string. The weight m1 acts upon the bucket m2 through the string and vice versa. According to Newton's third law of motion, it holds for the magnitudes of the string forces that:

    \[|\vec{T}_1| \,=\, |\vec{T}_2| \,=\, |\vec{T}|\,.\]

    We substitute into equations (1) and (2):

    \[m_1g - T \,=\, m_1a,\tag{3}\] \[T - m_2g \,=\, m_2a.\tag{4}\]

    Calculation of acceleration:

    By adding equations (3) and (4) we obtain:

    \[m_1g - m_2g \,=\, m_1a+m_2a,\] \[(m_1 - m_2)g \,=\, (m_1+m_2)a,\] \[a\,=\,\frac{(m_1 - m_2)g}{m_1+m_2}.\tag{5}\]

    We substitute numerically and calculate the acceleration:

    \[a\,=\,\left(\frac{\left(0.55\,-\,0.45\right)9.81}{0.55+0.45}\right)\,\mathrm{m\cdot s^{-2}}\,\dot=\,0.98\,\mathrm{m\cdot s^{-2}}\,.\]

    Tensile forces \(\vec{T}_1^\prime\) and \(\vec{T}_2^\prime\) act on both sides of the pulley. Also the force \(\vec{F}_\mathrm{p}\) with which the axis is fixed in place acts upon the pulley (we neglect the mass of the pulley and therefore the gravitational force acting upon it).

    Forces acting upon the pulley

    The translational acceleration is zero, so the resultant of the forces acting upon it is also zero:

    \[\vec{T}_1^\prime\,+\,\vec{T}_2^\prime\,+\,\vec{F}_\mathrm{p}\,=\,0\,.\tag{6}\]

    Because we neglect the mass of the pulley, it has no moment of inertia and does not affect the string forces. According to Newton's third law of motion, it holds for the string forces that:

    \[|\vec{T}_1^\prime| \,=\, |\vec{T}_2^\prime| \,=\, |\vec{T}|\,.\]

    We rewrite equation (6) using scalars:

    \[2T\,-\,F_\mathrm{p}\,=\,0,\] \[2T\,=\,F_\mathrm{p}.\tag{7}\]

    We can express the tensile force from equation (3) or (4):

    \[F_\mathrm{p}\,=\,2m_2\left(a+g\right)\,=\,2m_1\left(g-a\right)\,.\tag{8}\]

    We substitute numerically into equation (8):

    \[F_\mathrm{p}\,=\,2\,\cdot\,0.45\left(9.81\,+\,0.98\right)\,\mathrm{N}\,=\,2\,\cdot\,0.55\left(9.81\,-\,0.98\right)\,\mathrm{N},\] \[F_\mathrm{p}\,=\,0.9\,\cdot10.79\,\mathrm{N}\,=\,1.1\,\cdot8.83\,\mathrm{N},\] \[F_\mathrm{p}\,\dot=\,9.71\,\mathrm{N}.\]
  • ANSWER

    The acceleration of the system is \(a\,\dot=\,0.98\,\mathrm{m\cdot s^{-2}}\).

    The force with which the pulley acts upon its axis is \(F_\mathrm{p}\,\dot=\,9.71\,\mathrm{N}\).

Difficulty level: Level 2 – Upper secondary level
Tasks requiring deduction
Original source: Mandíková, D., Rojko, M.: Soubor úloh z mechaniky pro studium
učitelství. I. část. Interní materiál, MFF UK, Praha 1994 
Zpracováno v bakalářské práci Karolíny Slavíkové (2008).
×Original source: Mandíková, D., Rojko, M.: Soubor úloh z mechaniky pro studium učitelství. I. část. Interní materiál, MFF UK, Praha 1994
Zpracováno v bakalářské práci Karolíny Slavíkové (2008).
Send comment on task by email