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We can draw a diagram to show all the forces acting on each box and their accelerations. The surface is rough and friction will be limiting so equal to $\\mu R$.

Here, the mass of box $A$ is $m_1 = \\var{m1}\\mathrm{kg}$ and the mass of box $B$ is $m_2 = \\var{m2}\\mathrm{kg}$. The acceleration acts in the direction shown as box $A$ is heavier.

As the boxes are moving in different directions we can not model the whole system as a single particle.

a)

To find the normal reaction $R$ between box $B$ and the plane we resolve the forces perpendicular to the plane, where there is no acceleration.

\\begin{align}
R - m_2g \\cos \\theta & = 0, \\\\
R &= m_2 g \\cos \\theta, \\\\
& = \\var{m2} \\times 9.8 \\cos ( \\var{theta}{^\\circ}), \\\\
& = \\var{precround(R,3)} \\mathrm{N}.
\\end{align}

The normal reaction between the box $B$ and the plane is $\\var{precround(R,3)} \\mathrm{N}$.

b)

To find the acceleration we treat the boxes separately to get two equations involving the unknowns $T$ and $a$, then add them in order to cancel $T$.

Looking at box $B$ and resolving parallel to the plane in the direction of acceleration we have equation (1):

\\begin{align}
T - m_2g \\cos(90^{\\circ} - \\theta) - \\mu R & = m_2 a, \\\\
T - (\\var{m2} \\times g \\cos(\\var{90-theta}^{\\circ})) - (\\var{mu} \\times \\var{precround(R,3)}) & = \\var{m2}a, \\\\
\\simplify{T-{precround(t,3)}} &= \\var{m2}a.
\\end{align}

Looking at box $A$ and resolving in the direction of acceleration we have equation (2):

\\begin{align}
m_1g - T & = m_1a, \\\\
\\var{m1}g - T & = \\var{m1}a.
\\end{align}

Adding equations (1) and (2) gives

\\begin{align}
\\simplify[basic]{T - {precround(t,3)} +{m1}g -T} & = \\var{m2}a + \\var{m1}a, \\\\
\\var{m1}g - \\var{precround(m2*9.8*cos(radians(90-theta)) - mu*R,3)} & =(\\var{m2} + \\var{m1} )a, \\\\
\\var{precround(m1*9.8-m2*9.8*cos(radians(90-theta)) - mu*R,3)} & = \\var{m2+m1}a, \\\\
a & = \\var{precround((m1*9.8-m2*9.8*cos(radians(90-theta)) - mu*R)/(m2+m1),3)} \\mathrm{ms^{-2}}.
\\end{align}

The acceleration of the system is $\\var{precround(acceleration,3)} \\mathrm{ms^{-2}}$.

c)

We can find the tension in the string by substituting our value for acceleration into either equation (1) or (2).

This gives

\\begin{align}
\\var{m1}g - T & = \\var{m1}a, \\\\
T & = \\var{m1}g - \\var{m1}a, \\\\
& = \\var{m1} (9.8 - \\var{precround(acceleration,3)}), \\\\
& = \\var{precround(m1*(9.8 - acceleration),3)} \\mathrm{N}.
\\end{align}

The tension in the string is $\\var{precround(tension,3)} \\mathrm{N}.$

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Find the normal reaction force, $R$, between box $B$ and the plane. Give your answer in Newtons ($\\mathrm{N}$) to 3 decimal places.

$R = $ [[0]]

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The system is released from rest. Using your value of $R$ from part a) find the acceleration of the system, in $\\mathrm{ms^{-2}}$ to 3 decimal places.

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Using the answers to the previous parts, find the tension in the string, in Newtons ($\\mathrm{N}$) to 3 decimal places.

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You have not given your answer to the correct precision.

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A light inextensible string connects two boxes $A$ and $B$, of masses $m_1 = \\var{m1} \\mathrm{kg}$ and $m_2 = \\var{m2} \\mathrm{kg}$ respectively.

At the top of a rough inclined plane there is fixed a small smooth pulley over which the string passes. The plane is inclined to the horizontal at an angle $\\theta = \\var{theta}^{\\circ}$.

Box $A$ hangs on the edge of the plane with the string vertical and taut, whereas box $B$ rests on the inclined plane.

You are told that the coefficient of friction between $B$ and the plane is $\\mu = \\var{mu}$ and the acceleration due to gravity is $g = 9.8\\mathrm{ms^{-2}}$.

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An intermediary value to simplify the advice.

"}, "m2": {"definition": "random(0.5..5.5#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "m2", "description": ""}, "theta": {"definition": "random(10..70#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "theta", "description": ""}}, "metadata": {"description": "

Two particles connected by a string which passes over a pulley at the top of an inclined plane. Find the acceleration of the masses and the tension in the string. Can not model the whole system as a single particle.

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We can draw a diagram to show all the forces acting on each mass and the pulley, and the acceleration. Here, the mass of particle $A$ is $m_1 \\mathrm{kg} = \\var{A} \\mathrm{kg}$ and the mass of particle $m_2 \\mathrm{kg} = \\var{B} \\mathrm{kg}$. As particle $B$ is heavier the acceleration will act in the direction shown. The tension in the string is $T \\mathrm{N}$ and the force exerted on the string by the pulley is $F \\mathrm{N}$.

a) We can not treat the whole system as a particle because the particles are moving in different directions. Therefore to find the acceleration we need to resolve for each mass separately.

For $A$ we have equation (1): \\begin{align} T - m_1g & = m_1 a, \\\\
T - \\var{A}g & = \\var{A}a. \\end{align}

For $B$ we have equation (2): \\begin{align} m_2g - T & = m_2a, \\\\
\\var{B}g - T & = \\var{B}a. \\end{align}

In both (1) and (2) we have resolved in the direction of acceleration.

We can solve these two equations simultaneously. For example, by adding (1) and (2) we get

\\begin{align} T - \\var{A}g + \\var{B}g - T & = (\\var{A} + \\var{B}) a, \\\\
(\\var{B} - \\var{A})g & = (\\var{A} + \\var{B}) a, \\\\
a & = \\frac{\\var{A} + \\var{B}}{(\\var{B} - \\var{A})g}, \\\\
& = \\var{precround((B*g-A*g)/(A+B),3)} \\mathrm{ms^{-2}}. \\end{align}

The acceleration of each mass is $\\var{precround((B*g-A*g)/(A+B),3)} \\mathrm{ms^{-2}}$.

b) To find the tension in the string we can use our answer to part a) in either equation (1) or (2).

From equation (1) we have

\\begin{align} T - \\var{A}g & = \\var{A}a, \\\\
T & = \\var{A} \\times (g + a), \\\\
& = \\var{A} \\times (9.8 + \\var{ac}), \\\\
& = \\var{precround(A*(9.8+ac),3)} \\mathrm{N}. \\end{align}

The tension in the string is $\\var{T}\\mathrm{N}$.

c) The force exerted on the pulley by the string is $2T \\mathrm{N} = \\var{precround(2*T,3)} \\mathrm{N}$.

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Suppose that there are two particles $A$ and $B$, of masses $\\var{A} \\mathrm{kg}$ and $\\var{B} \\mathrm{kg}$ respectively. They are attached to a light inextensible string which passes over a small, smooth, fixed pulley. The masses hang with the string taut.

In the following answer to 3d.p. and take the acceleration due to gravity as $g = 9.8 \\mathrm{ms^{-2}}$.

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mass of particle A

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acceleration due to gravity

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mass of particle B

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acceleration

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Connected particle involving a pulley system - can not treat whole system as a whole due to different directions.

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Suppose that the system is released from rest. Find, in $\\mathrm{ms^{-2}}$, the acceleration of each mass.

Using your 3d.p. answer to part a) find the tension in Newtons ($\\mathrm{N}$) in the string.

Find the force in Newtons ($\\mathrm{N}$) exerted on the pulley by the string.

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Two masses in a box which is attached to a string. Finding acceleration by modelling the two masses as the whole system. Find forces exerted by the two masses.

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Suppose that there is a light box attached to a vertical light inextensible string. The box holds two masses $A$ and $B$ as shown in the diagram below, where $A$ rests upon $B$.

Suppose that the mass of $A$ is $m_1 = \\var{massa} \\, \\mathrm{kg}$ and the mass of $B$ is $m_2 = \\var{massb} \\, \\mathrm{kg}$. The acceleration due to gravity is $g=9.8 \\, \\mathrm{ms^{-2}}$.

Give your answers to the following questions to 3 decimal places.

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Suppose that the tension in the string is $\\var{T} \\, \\mathrm{N}$ and that the box is raised vertically, using the string. With what acceleration, in $\\mathrm{ms^{-2}}$, is the box raised?

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Using the acceleration in $\\mathrm{ms^{-2}}$ found in part a), find the force in Newtons ($\\mathrm{N}$) exerted on mass $B$ by mass $A$.

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Find the force in Newtons ($\\mathrm{N}$) exerted on mass $B$ by the box.

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acceleration 3d.p.

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tension in string

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a) We can draw a diagram to show the forces acting on the system. As all parts are moving in the same straight line (vertically) we can resolve for the whole system.

We have that

\\begin{align} T - m_1g - m_2g & = (m_1 + m_2)a, \\\\
\\var{T} - \\var{massa}g - \\var{massb}g & = \\var{mass}a, \\\\
a & =\\frac{\\var{T} - \\var{massa}g - \\var{massb}g}{\\var{mass}}, \\\\
& = \\var{a}\\mathrm{ms^{-2}}.\\end{align}

The acceleration of the system is $\\var{a}\\mathrm{ms^{-2}}$.

b) To find the force exerted on mass $B$ by mass $A$ we can find the force exerted on $A$ by $B$ (the normal reaction, $R$) and use Newton's 3rd Law to say that the force exerted on $B$ by $A$ will have the same magnitude.

We resolve the forces to find $R$, using the 3d.p. value for $a$ from part a).

\\begin{align} R - m_1g & = m_1a, \\\\
R & = m_1g + m_1a, \\\\
& = \\var{massa} \\left(9.8 + \\var{precround(aexact,4)}\\right), \\\\
& = \\var{precround(massa*(9.8+aexact),3)} \\mathrm{N}.\\end{align}

So the force exerted on $B$ by $A$ is $\\var{precround(massa*(9.8+aexact),3)} \\mathrm{N}$.

c) Let X be the force exerted on mass $B$ by the box.

Resolving forces acting on mass $B$, with upwards being the positive direction, and using our answer from (b), we obtain:

\\begin{align} X - \\var{precround(massa*(9.8+aexact),4)} - m_2g & = m_2a, \\\\
X & = \\var{precround(massa*(9.8+aexact),4)}+\\var{massb}g+\\var{massb}\\times \\var{precround(aexact,4)}, \\\\
X & = \\var{precround(massa*(9.8+aexact)+massb*g+massb*aexact,3)} \\mathrm{N} \\end{align}

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To be used on the connected particles page under the Dynamics section of the Mechanics wiki page.

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