Collection of Solved Problems in Physics

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?

• 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?

• Solution of hint 1

  

  Weight P:

  • \(\vec{F}_{Gp}\) .......weight force (F_Gp = m_pg)

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

  Weight B:

  • \(\vec{F}_{Gb}\) .......weight force (F_Gb = m_bg)

  • \(\vec{F^,}_{p}\) ....... 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 F_p = 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.

  Závaží P:

  \[\vec{F}_{Gp}+\vec{F}_{p}=\vec{0}\,.\tag{1}\]

  Závaží B:

  \[\vec{F}_{Gb}+\vec{F'}_{p}+\vec{F}_{a}=\vec{0}\,.\tag{2}\]

• 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 of hint 2

  

  Weight P:

  \[{F}_{p}=F_{Gp}=m_pg\,.\]

  Weight B:

  \[x:\hspace{60px}F_p\sin\beta-F_a\sin\alpha=0\tag{3}\]

  \[y:\hspace{10px}F_a\cos\alpha+F_p\cos\beta-F_{Gb}=0\tag{4}\]

  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:

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

  So:

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

  We substitute gained F_p 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, m_b:

  \[m_p=10\frac {\sin45^\circ}{\sin(45^\circ+60^\circ)}\,\mathrm{kg}\dot{=}7.3 \,\mathrm{kg}\,,\] \[F_a=10{\cdot}10\frac {\sin60^\circ}{\sin(45^\circ+60^\circ)}\,\mathrm{N} \dot{=} 90 \,\mathrm{N}\,.\]

• Solution

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

  

  Weight P:

  • \(\vec{F}_{Gp}\) .......weight force (F_Gp = m_pg)

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

  Weight B:

  • \(\vec{F}_{Gb}\) .......weight force (F_Gb = m_bg)

  • \(\vec{F^,}_{p}\) ....... 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 F_p = 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:

  \[\vec{F}_{Gp}+\vec{F}_{p}=\vec{0}\,.\tag{1}\]

  Weight B:

  \[\vec{F}_{Gb}+\vec{F'}_{p}+\vec{F}_{a}=\vec{0}\,.\tag{2}\]

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

  Weight P:

  \[{F}_{p}=F_{Gp}=m_pg\,.\]

  Weight B:

  \[x:\hspace{60px}F_p\sin\beta-F_a\sin\alpha=0\tag{3}\]

  \[y:\hspace{10px}F_a\cos\alpha+F_p\cos\beta-F_{Gb}=0\tag{4}\]

  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:

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

  So:

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

  We substitute gained F_p 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, m_b:

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

• Answer

  Demanded mass of weight P is equal to:

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

  The stress in rope AB is equal to:

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

  Numerically:

  \[m_p \dot{=}7.3 \,\mathrm{kg}\,,\] \[F_a \dot{=} 90 \,\mathrm{N}\,.\]