## Cubes banded by fibers

### Task number: 1524

Three cubes weighing *m*_{1}, *m*_{2} and *m*_{3} are mutually banded by fibers, as shown in the igure. The coefficient of friction between all contact surfaces is identical and has a value of *f*. Determine the acceleration of cubes and tensile force tensioning both fibres. Do not consider the weight of the pulleys and the fiber.

#### Notation

*m*_{1}mass of cube number one *m*_{2}mass of cube number two *m*_{3}mass of cube number three *f*friction coefficient for every arbitrary contact *a*= ?acceleration of cubes *T*_{1}= ?tensile force stretching the fibre between cubes *m*_{2}and*m*_{3}*T*_{2}= ?tensile force stretching the fibre between cubes *m*_{3}and*m*_{1}#### Hint 1 – forces acting on the individual cubes, motion equations

Be aware of the forces acting on the individual cubes and type equation of motion for each of them.

**Recommendation:**For clarity, draw a figure in which you will draw the forces acting on the cubes. Draw the figures for each of the cubes separately.#### Hint 2 – scalar equations of motion, the relationship among tensile forces in fibers, frictional forces and pressure forces

Choose the appropriate coordinate system and rewrite the scalar equations of motion. Recognize what relationships are among some tensile, frictional and pressure forces – Use 3rd Newton’s law.

#### Hint 3 – frictional forces

Be aware of what the value of frictional force depends on and how it can be expressed. Quantify frictional forces \(F_{t_2}\) and \(F_{t_3}\).

#### Hint 4 – Calculation of acceleration and tensile forces in fibres

Use equation (1), (2) and (6). After substituting for frictional forces you get three equations with three unknowns. Calculate the values of acceleration and tension forces

*T*_{1}and*T*_{2}in fibres.#### TOTAL SOLUTION

We draw the forces acting on the cubes to the figure at first and write the equations of motion for them.

**The forces acting on the cubes:**Cube

*m*_{1}:\(\vec{F}_{G_1}\)…weight force

\(\vec{T}_{1}\)…tensile force the fibre is acting on the cube

Cube

*m*_{2}:\(\vec{F}_{G_2}\)…weight force

\(\vec{N}_{2}\)…compressive force which the cube

*m*_{2}pushes the cube*m*_{3}with\(\vec{T}_{2}\)…tensile force the fibre is acting on the cube with

\(\vec{F}_{t_2}\)…frictional force between the cubes

*m*_{2}and*m*_{3}Cube

*m*_{3}:\(\vec{F}_{G_3}\)…weight force

\(\vec{T}_{2}\prime\)…tensile force the fibre is acting on the cube

\(\vec{T}_{1}\prime\)…tensile force the fibre is acting on the cube

\(\vec{F}_{t_3}\)…frictional force between the cube and the ground

\(\vec{F}_{t_2}\prime\)…frictional force between the cubes

*m*_{2}and*m*_{3}\(\vec{N}_{2}\prime\)…compressive force which the cube

*m*_{3}pushes cube*m*_{2}with\(\vec{N}_{3}\)…compressive force which the cube the ground with

**The equations of motion for each cube:**All three cubes are interconnected by fibres, moving with the same acceleration

\[m_1:\qquad \vec{F}_{G_1} + \vec{T}_1=m_1\vec{a}\] \[m_2:\qquad \vec{F}_{G_2}+\vec{N}_2+\vec{F}_{t_2}+ \vec{T}_2=m_2\vec{a}\] \[m_3:\qquad \vec{F}_{G_3}+\vec{T}_1\prime +\vec{T}_2\prime+\vec{N}_3+\vec{N}_2\prime+\vec{F}_{t_3}+\vec{F}_{t_2}\prime=m_3\vec{a}\]To rewrite the equations of motion as a scalar equation, we choose a coordinate system

*x*,*y*so that the*x*-axis goes along the direction of movement of the cubes. The*y*-axis is perpendicular to the*x*-axis.**Scalar equation of motion:**Cube

\[x:\qquad F_{G_1} - T_1\,=\,m_1a\tag{1}\]*m*_{1}:Cube

\[x:\qquad T_2-F_{t_2}\,=\,m_2a\tag{2}\] \[y:\qquad N_2-F_{G_2}\,=\,0\tag{3}\]*m*_{2}:Cube

\[x:\qquad T_1\prime-T_2\prime-F_{t_3}-F_{t_2}\prime\,=\,m_3a\tag{4}\] \[y:\qquad N_3-N_2\prime-F_{G_3}\,=\,0\tag{5}\]*m*_{3}:The cubes do not move in the

*y*direction, so the acceleration is zero.**Tensile forces in the fibres:**Since we neglect the weight of the pulleys, thus they have no inertia and do not affect the tensile forces in fibres. Cube

\[|\vec{T}_1| \,=\, |\vec{T}_1\prime|\] \[|\vec{T}_2| \,=\, |\vec{T}_2\prime|\]*m*_{3}acts cube*m*_{2}through the fibre and contrariwise cube*m*_{2}acts on cube*m*_{3}. Similarly, with the cube*m*_{1}acts on cube*m*_{3}. According to the Newton’s third Law it applies for the value of the forces:**Frictional forces:**Frictional force of cube

*m*_{3}acts on cube*m*_{2}and contrariwise frictional force of cube*m*_{2}acts on cube*m*_{3}in the opposite direction. It applies according to the Newton’s third Law:\[|\vec{F}_{t_2}| \,=\, |\vec{F}_{t_2}\prime|\]

**Compressive forces:**Compressive force of cube

\[|\vec{N}_2| \,=\, |\vec{N}_2\prime|\]*m*_{3}acts on cube*m*_{2}and contrariwise compressive force of cube*m*_{2}acts on cube*m*_{3}in the opposite direction. It applies according to the Newton’s third Law:We rewrite the equations (4) and (5):

\[x:\qquad T_1-T_2-F_{t_3}-F_{t_2}\,=\,m_3a\tag{6}\] \[y:\qquad N_3-N_2-F_{G_3}\,=\,0\tag{7}\]**Expression of friction forces:**Frictional force \(F_{t_2}\) is proportional to the force which cube

*m*_{2}pushes cube*m*_{3}with, that is, according to Newton’s third law as large, as the force which the cube*m*_{3}pushes cube*m*_{2}with.Frictional force \(F_{t_2}\) can be expressed as:

\[F_{t_2} \,=\, N_2f\]Substituting for \(N_2\) from the equation (3):

\[F_{t_2} \,=\, F_{G_2}f\,=\,m_2gf\tag{8}\]Frictional force \(F_{t_3}\) is proportional to the force which cube

*m*_{3}affects the table with, that is, according to Newton’s third law as large, as the force which the table affects cube*m*_{3}.Frictional force \(F_{t_3}\) can be expressed as:

\[F_{t_3} \,=\, N_3f\]

Substituting for \(N_3\) from the equation (7):

\[F_{t_3} \,=\, (N_2+F_{G_3)}f\,=\,(F_{G_2}+F_{G_3})f\] \[F_{t_3} \,=\, (m_2+m_3)gf\tag{9}\]We rewrite the equation (1), (2) and (6) into equation (2), substitute the frictional force \(F_{t_2}\) and to the equation (6) substitute the frictional forces \(F_{t_2}\) and \(F_{t_3}\):

\[m_1g - T_1\,=\,m_1a\tag{10}\] \[T_2-m_2gf\,=\,m_2a\tag{11}\] \[T_1-T_2-(m_2+m_3)gf-m_2gf\,=\,m_3a\tag{12}\]**Calculation of acceleration:**We count the equations (10) – (12) up.

\[m_1g-m_2gf-m_3gf-2m_2gf\,=\,\left(m_3+m_2+m_1\right)a\] \[m_1g-m_3gf-3m_2gf\,=\,\left(m_3+m_2+m_1\right)a\]\[a\,=\,\frac{g\left(m_1-m_3f-3m_2f\right)}{m_3+m_2+m_1}\tag{13}\]

**Calculation of tensile force***T*_{1}:We express the force

*T*_{1}from equation (10):\[T_1\,=\,m_1g-m_1a\,=\,m_1g-m_1\frac{m_1g-m_3gf-3m_2gf}{m_3+m_2+m_1}\]

We convert it to the last common denominator:

\[T_1\,=\,\frac{m_1^2g+m_1m_2g+m_1m_3g-m_1^2g+m_1m_3gf+3m_1m_2gf} {m_3+m_2+m_1}\] \[T_1\,=\,\frac{m_1m_3gf+3m_1m_2gf+m_1m_2g+m_1m_3g} {m_3+m_2+m_1}\] \[T_1\,=\,\frac{m_1g\left[m_3(f+1)+ m_2(3f+1)\right]} {m_3+m_2+m_1}\tag{14}\]**Calculation of tensile force***T*_{2}:We express the force

\[T_2\,=\,m_2a\,+\,m_2gf\,=\,m_2\frac{m_1g-m_3gf-3m_2gf} {m_3+m_2+m_1}\,+\,m_2gf\]*T*_{2}from equation (11):We convert it to the last common denominator:

\[T_2\,=\,\frac{m_1m_2g-m_2m_3gf-3m_2^2gf+m_1m_2gf+m_2^2gf+m_2m_3gf} {m_3+m_2+m_1}\] \[T_2\,=\,\frac{m_1m_2g-2m_2^2gf+m_1m_2gf} {m_3+m_2+m_1}\] \[T_2\,=\,\frac{m_2g\left[m_1(1+f)-2m_2f\right]} {m_3+m_2+m_1}\tag{15}\]#### ANSWER

The value of acceleration of the cubes is: \[a\,=\,\frac{g\left(m_1-m_3f-3m_2f\right)}{m_3+m_2+m_1}.\]

The value of the tensile force stretching the fibre between the cubes

\[T_1\,=\,\frac{m_1g\left[m_3\left(f+1\right)+ m_2\left(3f+1\right)\right]}{m_3+m_2+m_1}\,.\]*m*_{2}and*m*_{3}isThe value of the tensile force stretching the fibre between the cubes

\[T_2\,=\,\frac{m_2g\left[m_1\left(1+f\right)-2m_2f\right]}{m_3+m_2+m_1}\,.\]*m*_{3}and*m*_{1}is