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6.2: Solving Problems with Newton's Laws (Part 1)

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  • This figure shows the development of the free body diagram of a piano being lifted and passed through a window. Figure a is a sketch showing the piano hanging from a crane and part way through a window. Figure b identifies the forces. It shows the same sketch with the addition of the forces, represented as labeled vector arrows. Vector T points up, vector F sub T points down, vector w points down. Figure c defines the system of interest. The sketch is  shown again with the piano circled and identified as the system of interest. Only vectors T up and w down are included in this diagram. The downward force F sub T is not a force on the system of interest since it is exerted on the outside world. It must be omitted from the free body diagram. The free body diagram is shown as well. It consists of a dot, representing the system of interest, and the vectors T pointing up and w pointing down, with their tails at the dot. Figure d shows the addition of the forces. Vectors T and w are shown. We are told that these forces must be equal and opposite since the net external force is zero. Thus T is equal to minus w.
    Figure \(\PageIndex{6}\): An Atwood machine and free-body diagrams for each of the two blocks.


    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 a2 in the downward direction, block 1 accelerates upward with acceleration a1. Thus, a = a1 = −a2.


    1. We have $$For\; m_{1}, \sum F_{y} = T − m_{1}g = m_{1}a \ldotp \quad For\; m_{2}, \sum F_{y} = T − m_{2}g = −m_{2}a \ldotp$$(The negative sign in front of m2 a indicates that m2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is $$(m_{2} - m_{1})g = (m_{1} + m_{2})a \ldotp$$Solving for a: $$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}}g = \frac{4\; kg - 2\; kg}{4\; kg + 2\; kg} (9.8\; m/s^{2}) = 3.27\; m/s^{2} \ldotp$$
    2. Observing the first block, we see that $$T − m_{1}g = m_{1}a$$ $$T = m_{1}(g + a) = (2\; kg)(9.8\; m/s^{2} + 3.27\; m/s^{2}) = 26.1\; N \ldotp$$


    The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, (m2 − m1)g, to the total mass of the system, m1 + m2. We can also use the Atwood machine to measure local gravitational field strength.

    Exercise 6.3

    Determine a general formula in terms of m1, m2 and g for calculating the tension in the string for the Atwood machine shown above.


    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).