Weight on rope

Task number: 1521

Weight B with mass of 10 kg is fastened to a rope, as shown in the Figure. The distance AB is constant. How heavy weight P is it necessary to choose for α = 45°, β = 60°? How big will the stress be in rope AB in this case?
zadání 521
  • Notation

    mb = 10 kg mass of weight B
    α = 45° desired angle between rope AB and the vertical line
    β = 60° desired angle between the rope from B to the pulley and the vertical line
    mp =? (kg) mass of weight P
    Fa =? (N) stress of rope AB
  • Hint 1

    Draw all the forces acting on the weights B and P in the figure. What must be valid for their resultant?
  • Hint 2

    Choose an appropriate coordinate system and rewrite the equations (1) and (2) as a scalar. Express the searched parameters from the obtained equations.

  • Solution

    We draw all the forces acting on the weights B and P to the figure and choose the coordinate system.

    Popis obrázku

    Weight P:

    • \(\vec{F}_{Gp}\) .......weight force (FGp = mpg)

    • \(\vec{F}_{p}\).......force, by which the rope is pulled

    Weight B:

    • \(\vec{F}_{Gb}\) .......weight force (FGb = mbg)

    • \(\vec{F^,}_{p}\) ....... force, which the rope with the weight P is pulled by

      (The pulley is at standstill so that it is valid for the values of forces \(\vec{F}_{p}\) and \(\vec{F^,}_{p}\) that Fp = F´p, The weight P pulls the rope with the weight B and vice versa)

    • \(\vec{F}_{a}\) ....... force, by which the rope mounted to the wall is pulled

    Both weights are at standstill so the resultant of acting forces must be equal to the zero vector.

    Weight P:


    Weight B:


    We rewrite the equations (1) and (2) as a scalar:

    Weight P:


    Weight B:



    From equation (3):

    \[F_a = F_p \frac{\sin \beta}{\sin \alpha}\,.\tag{5}\]

    We substitute into the equation (4):

    \[F_p \frac{\sin \beta}{\sin \alpha} \cos \alpha + F_p \cos \beta = F_{Gb}\,.\]

    And modify:

    \[F_p\frac{\sin\beta\cos\alpha+\sin\alpha\cos \beta}{\sin \alpha} = F_{Gb}\,.\]

    Using the addition formula for sine:

    \[F_p\sin(\alpha+\beta)=F_{Gb}\sin\alpha\,,\] \[F_p=F_{Gb}\frac{\sin\alpha}{\sin(\alpha+\beta)}\,.\]


    \[gm_p=gm_b\frac{\sin\alpha}{\sin(\alpha+\beta)}\,,\] \[m_p=m_b\frac{\sin\alpha}{\sin(\alpha+\beta)}\,.\]

    We substitute gained Fp to the equation (5):

    \[F_a = F_{Gb} \frac{\sin \alpha}{\sin (\alpha + \beta)} \frac{\sin \beta}{\sin \alpha}\,,\] \[F_a = F_{Gb} \frac{\sin \beta}{\sin (\alpha + \beta)}\,,\] \[F_a = gm_{b} \frac{\sin \beta}{\sin (\alpha + \beta)}\,.\]

    We substitute given values:

    α, β, mb:

    \[m_p=10\frac {\sin45^\circ}{\sin(45^\circ+60^\circ)}\,\mathrm{kg}\dot{=}7{,}3 \,\mathrm{kg}\,,\] \[F_a=10{\cdot}10\frac {\sin60^\circ}{\sin(45^\circ+60^\circ)}\,\mathrm{N} \dot{=} 90 \,\mathrm{N}\,.\]
  • Answer

    Demanded mass of weight P is equal to:


    The stress in rope AB is equal to:

    \[F_a = gm_{b} \frac{\sin \beta}{\sin (\alpha + \beta)} \,.\]


    \[m_p \dot{=}7.3 \,\mathrm{kg}\,,\] \[F_a \dot{=} 90 \,\mathrm{N}\,.\]
Difficulty level: Level 2 – Upper secondary level
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