Chapter 7: Q.58 (page 181)

$F I G U R E C P 7.58$ shows three hanging masses connected by massless strings over two massless, frictionless pulleys.
a. Find the acceleration constraint for this system. It is a single equation relatinglocalid="1649865348893" $a_{1 y}, a_{2 y}, a n d a_{3 y} .$ Hint: $y_{A}$ isn’t constant.
b. Find an expression for the tension in string $A .$ Hint: You should be able to write four second-law equations. These, plus the acceleration constraint, are five equations in five unknowns.
c. Suppose: $m_{1} = 2.5 k g, m_{2} = 1.5 k g, a n d m_{3} = 4.0 k g .$ Find the acceleration of each.
d. The $4.0 k g$ mass would appear to be in equilibrium. Explain why it accelerates.

Short Answer

Expert verified

a. the acceleration equation is given by $a_{1 y} + a_{2 y} + 2 a_{3 y} = 0$

b. the tension equation for pulley A is given by $2 T = (m_{1} + m_{2} + m_{3}) g - (m_{1} a_{1 y} + m_{2} a_{2 y} + m_{3} a_{3 y})$

c. $a_{1 y} = a_{2 y} = a_{3 y} = 9.8 m / s^{2}$

d. Yes. The $4.0 k g$ mass would appear to be in equilibrium.

Step by step solution

Explanation (Part a)

Since the strings and pulleys are massless, evaluate the displacement with respect to pulley $A$ and $B$ which is shown in the $F I G U R E C P 7.58$

For pulley $A :$ $y_{1} + y_{2} - 2 y_{A} = 0 (I)$

For pulley $B :$ $y_{A} + y_{3} = 0 (I I)$

multiply equation $(I I)$ by $2$ and add it to equation $(I)$

Therefore, the displacement equation becomes, $y_{1} + y_{2} + 2 y_{3} = 0$

differentiating double the times to yield the acceleration equation.

${y_{1}}^{''} + {y_{2}}^{''} + {2 y_{3}}^{''} = 0 a_{1 y} + a_{2 y} + 2 a_{3 y} = 0$

Explanation (Part b)

Newton's second law equation for mass $m_{1}$

$- T_{1} + m_{1} g = m_{1} a_{1 y}$

Newton's second law equation for mass $m_{2}$

$- T_{2} + m_{2} g = m_{2} a_{2 y}$

Newton's second law equation for mass $m_{1}$ and $m_{2}$

$(m_{1} + m_{2}) g - T_{3} = m_{1} a_{1 y} + m_{2} a_{2 y}$

Newton's second law equation for mass $m_{3}$

$m_{3} g - T_{4} = m_{3} a_{3 Y}$

Since all the pulleys are massless, $T_{1} = T_{2} a n d T_{3} = T_{4}$

Also we know that acceleration equation as follows: $a_{1 y} + a_{2 y} + 2 a_{3 y} = 0$

Adding first two equations of pulley $A$

$- 2 T + (m_{1} + m_{2}) g = m_{1} a_{1 y} + m_{2} a_{2 y}$

Simplifying the next two equation gives

$- 2 T + (m_{1} + m_{2} + m_{3}) g = m_{1} a_{1 y} + m_{2} a_{2 y} + m_{3} a_{3 y} 2 T = (m_{1} + m_{2} + m_{3}) g - (m_{1} a_{1 y} + m_{2} a_{2 y} + m_{3} a_{3 y})$

Explanation (part c)

Using tension equation we got,

$2 T = (m_{1} + m_{2} + m_{3}) g - (m_{1} a_{1 y} + m_{2} a_{2 y} + m_{3} a_{3 y}) 2 T = 78.4 - (2.5 a_{1 y} + 1.5 a_{2 y} + 4 a_{3 y}) 39.2 - (2.5 a_{1 y} + 1.5 a_{2 y}) = 78.4 - (2.5 a_{1 y} + 1.5 a_{2 y} + 4 a_{3 y}) 4 a_{3 y} = 39.2 a_{3 y} = 9.8 m / s^{2}$

Therefore the acceleration equation becomes

$a_{1 y} + a_{2 y} = 19.6 m / s^{2}$

Explanation (Part d)

Yes. The $4.0 k g$ mass would appear to be in equilibrium. Because the acceleration of each mass is said to be equal and follow the acceleration in its path due to massless pulley and strings and frictionless.