# What is the acceleration of the system and the tension in the ropes joining the masses in the following diagram?

Nov 17, 2016

To 3sf we have

$a = 0.191 m {s}^{-} 2$
${T}_{1} = 103 N$
${T}_{2} = 111 N$

#### Explanation:

Assumptions
In order to answer this question we have to make a few assumptions.
1) The ropes are non-elastics (As we assume the tension in the rope is the same either side of the pulleys)
2) The pulleys are smooth (so that is no friction between the rope and the pulleys)
3) $g = 9.8 m {s}^{-} 2$

Laying Out The Problem
The forces have been drawn on the diagram, I will use the symbols $A$,$B$ and $C$ to represent the 12Kg, 7Kg as 23Kg objects respectively, along with subscripts for the coefficients of friction, $\mu$, and reactions forces $R$

It should be clear this system will not stay in equilibrium, rather objects $A$,$B$ and $C$ will start to accelerate under gravity to the right. All objects will experience the same constant acceleration, $a$.

Reaction Forces
The first thing we can do is calculate the reaction forces for each of the objects. We need to do this s the friction force is related to the reaction, by $F = \mu R$

If we apply NSL to A perpendicular to the slope we get;

$12 g \cos 55 - {R}_{A} = 0 \implies {R}_{A} = 12 g \cos 55$

Similarly for B and C we get:

${R}_{B} = 7 g$
${R}_{C} = 23 g \cos 32$

Consider Object A
If we apply NSL upwards parallel to the slope we get;

${T}_{1} - {F}_{A} - 12 g \sin 55 = 12 a$
$\therefore {T}_{1} - {\mu}_{A} {R}_{A} - 12 g \sin 55 = 12 a$
$\therefore {T}_{1} - \left(0.07\right) \left(12 g \cos 55\right) - 12 g \sin 55 = 12 a$
$\therefore {T}_{1} - 0.84 g \cos 55 - 12 g \sin 55 = 12 a$
$\therefore {T}_{1} = 12 a + 0.84 g \cos 55 + 12 g \sin 55$ ..... [1]

Consider Object C
If we apply NSL downwards parallel to the slope we get;

$23 g \sin 32 - {T}_{2} - {F}_{C} = 23 a$
$\therefore 23 g \sin 32 - {T}_{2} - {\mu}_{C} {R}_{C} = 23 a$
$\therefore 23 g \sin 32 - {T}_{2} - \left(0.04\right) \left(23 g \cos 32\right) = 23 a$
$\therefore 23 g \sin 32 - {T}_{2} - 0.92 g \cos 32 = 23 a$
$\therefore {T}_{2} = 23 g \sin 32 - 0.92 g \cos 32 - 23 a$ ..... [2]

Consider Object B
If we apply NSL parallel to the slope we get;

${T}_{2} - {T}_{1} - {F}_{B} = 7 a$
${T}_{2} - {T}_{1} - {\mu}_{B} {R}_{B} = 7 a$
${T}_{2} - {T}_{1} - \left(0.1\right) \left(7 g\right) = 7 a$
${T}_{2} - {T}_{1} - 0.7 g = 7 a$ ..... [3]

Now if we substitute [1] and [2] into [3] we get:

$23 g \sin 32 - 0.92 g \cos 32 - 23 a - \left(12 a + 0.84 g \cos 55 + 12 g \sin 55\right) - 0.7 g = 7 a$
$\therefore 23 g \sin 32 - 0.92 g \cos 32 - 23 a - 12 a - 0.84 g \cos 55 - 12 g \sin 55 - 0.7 g = 7 a$
$\therefore 42 a = 23 g \sin 32 - 0.92 g \cos 32 - 0.84 g \cos 55 - 12 g \sin 55 - 0.7 g$
$\therefore a = \frac{g}{42} \left\{23 \sin 32 - 0.92 \cos 32 - 0.84 \cos 55 - 12 \sin 55 - 0.7\right)$
$\therefore a = 0.191339 \ldots m {s}^{-} 2$

From [1] we have

${T}_{1} = 12 \left(0.191339 \ldots\right) + 0.84 g \cos 55 + 12 g \sin 55$
$\therefore {T}_{1} = 103.350036 \ldots$

From [2] we have;

${T}_{2} = 23 g \sin 32 - 0.92 g \cos 32 - 23 \left(0.191339 \ldots\right)$
$\therefore {T}_{2} = 23 g \sin 32 - 0.92 g \cos 32 - 23 \left(0.191339 \ldots\right)$
$\therefore {T}_{2} = 111.453731 \ldots$

Hence, to 3sf we have

$\therefore a = 0.191 m {s}^{-} 2$
$\therefore {T}_{1} = 103 N$
$\therefore {T}_{2} = 111 N$