6.1 Solving problems with newton’s laws (Page 7/12)

University physics volume 1 Page 7 / 12

For block 1: $\vec{T} + {\vec{w}}_{1} + \vec{N} = m_{1} {\vec{a}}_{1}$

For block 2: $\vec{T} + {\vec{w}}_{2} = m_{2} {\vec{a}}_{2} .$

Notice that $\vec{T}$ is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

Solution

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

\begin{matrix} Block 1 & Block 2 \\ \sum F_{x} = m a_{x} & \sum F_{y} = m a_{y} \\ T_{x} = m_{1} a_{1 x} & T_{y} - m_{2} g = m_{2} a_{2 y} . \end{matrix}

When block 1 moves to the right, block 2 travels an equal distance downward; thus, $a_{1 x} = − a_{2 y} .$ Writing the common acceleration of the blocks as $a = a_{1 x} = − a_{2 y},$ we now have

T = m_{1} a

and

T - m_{2} g = − m_{2} a .

From these two equations, we can express a and T in terms of the masses $m_{1} and m_{2}, and g :$

a = \frac{m_{2}}{m_{1} + m_{2}} g

and

T = \frac{m_{1} m_{2}}{m_{1} + m_{2}} g .

Significance

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set

T = m_{2} g .

You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Check Your Understanding Calculate the acceleration of the system, and the tension in the string, when the masses are $m_{1} = 5.00 kg$ and $m_{2} = 3.00 kg .$

$a = 3.68 {m/s}^{2},$ $T = 18.4 N$

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Atwood machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine , which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In [link] ,

m_{1} = 2.00 kg

and

m_{2} = 4.00 kg .

Consider the pulley to be frictionless. (a) If

m_{2}

is released, what will its acceleration be? (b) What is the tension in the string?

An Atwood machine consists of masses suspended on either side of a pulley by a string passing over the pulley. In the figure, mass m sub 1 is on the left and mass m sub 2 is on the right. The free body diagram of block one shows mass one with force vector T pointing vertically up and force vector w sub one pointing vertically down. The free body diagram of block two shows mass two with force vector T pointing vertically up and force vector w sub two pointing vertically down. — An Atwood machine and free-body diagrams for each of the two blocks.

Strategy

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration

a_{2}

in the downward direction, block 1 accelerates upward with acceleration

a_{1}

. Thus,

a = a_{1} = − a_{2} .

Solution

We have

$\begin{matrix} For m_{1}, \sum F_{y} = T - m_{1} g = m_{1} a . & For m_{2}, \sum F_{y} = T - m_{2} g = − m_{2} a . \end{matrix}$

(The negative sign in front of $m_{2} a$ indicates that $m_{2}$ accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

$(m_{2} - m_{1}) g = (m_{1} + m_{2}) a .$

Solving for a :

$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g = \frac{4 kg - 2 kg}{4 kg + 2 kg} (9.8 {m/s}^{2}) = 3.27 {m/s}^{2} .$
Observing the first block, we see that

$\begin{matrix} T - m_{1} g = m_{1} a \\ T = m_{1} (g + a) = (2 kg) (9.8 {m/s}^{2} + 3.27 {m/s}^{2}) = 26.1 N . \end{matrix}$

Significance

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system,

(m_{2} - m_{1}) g

, to the total mass of the system,

m_{1} + m_{2}

. We can also use the Atwood machine to measure local gravitational field strength.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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