5.6: Non-Inertial Frame of Reference and Inertial Forces (2023)

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    In the previous sections, we described how to use Newton's First Law to identify an inertial frame of reference (one in which Newton's First Law is true) to identify the forces exerted on an object so that Newton's Second Law can be applied. It is possible to apply Newton's Laws in a non-inertial frame of reference,provided an additional “inertial force” is included.

    Suppose we hang a mass,\(subway\), from the roof of our car with a rope. If the car accelerates forward with constant acceleration,\(\what\), the mass will swing towards the rear of the car and the rope will not stay vertical while the car maintains its constant acceleration, as shown in Figure \(\PageIndex{1}\). As long as the car maintains its acceleration, the sprung mass will not move relative to the car.

    5.6: Non-Inertial Frame of Reference and Inertial Forces (2)

    We can analyze this movement from the inertial frame of reference of the ground. In this frame of reference, there are two forces exerted on the mass:

    1. \(\vec F_g\), its weight, with magnitude\(mg\).
    2. \(\what\), a pulling force exerted by the rope, in the direction of the rope.

    The two forces are shown in the free-body diagram of Figure \(\PageIndex{2}\), along with a coordinate system chosen such that\(X\)points in the direction of the acceleration of the mass (which is the same as the acceleration of the car, since the mass is not moving relative to the car).

    5.6: Non-Inertial Frame of Reference and Inertial Forces (3)

    writing the\(X\)y\(y\)components of Newton's Second Law for mass, we have: \[\begin{aligned} \sum \vec F &= \vec T + \vec F_g= m \vec a\\ \therefore\sum F_x &= T\ sin \ theta = ma\\ \so\sum F_y &= T\cos\theta-F_g=0\end{aligned}\] Instead, we can model the motion of the mass in the car's frame of reference by pretending we're sitting in the car. In the car's frame of reference, the mass is stationary and therefore has no acceleration. In the car's non-inertial frame of reference, we still have the weight and pull forces exerted on the mass; these have the same magnitude and direction as in the ground inertial frame. You could replace the string with a spring scale, and observers in the car and on the ground would agree that the spring scale reads the same number. These observers also agree that the weight of the dough is the same. However, the two observers disagree on whether the mass is accelerating, as the observer in the car measures that the mass has no acceleration.

    In the car's frame of reference, the acceleration of the mass is zero. If we want Newton's Second Law to hold, it implies that, in the car's frame of reference, the sum of the forces on the mass must be zero: \[\begin{aligned} \sum \vec F & = 0 \quad \ quad \text{(car's frame of reference)}\end{aligned}\] By analyzing the motion of the ground frame, we know that the vector sum of the forces\(\what\)y\(\vec F_g\)It's the same as\(m\what\). The only way the force on the car's frame of reference adds to zero is if there is an additional force,\(\what F_I\), which is exercised in this frame: \[\begin{aligned} \sum \vec F &= \vec T + \vec F_g + \vec F_I =0\quad\quad\text{(car frame )} \end{ aligned}\] As we know that\(\vec T + \vec F_g=m\vec a\), we can plug that into the equation above: \[\begin{aligned} \sum \vec F &= \vec T + \vec F_g + \vec F_I =0\quad\quad\text{(car reference) } \\ &=m\vec a+\vec F_I = 0\\ \so F_I &= -m\vec a\end{aligned}\] and we find that this “inertial force”,\(\what F_I\), must be exerted in the opposite direction to the acceleration of the reference frame, with module given by\(breasts\). The free-body diagram for the mass, as seen in the car's frame of reference, is illustrated in figure \(\PageIndex{3}\).

    5.6: Non-Inertial Frame of Reference and Inertial Forces (4)

    Example \(\PageIndex{1}\)

    You are in an elevator that descends with constant acceleration.\(\what\). You are standing on a spring scale. What is the value of its weight as shown on the spring scale? Assume your mass is\(subway\). (The spring scale will display your weight with the same magnitude as the normal force the scale exerts on you.)


    We can model its motion in the elevator's non-inertial frame, where its acceleration is zero. The forces exerted on you are:

    1. \(\vec F_g\), its weight, with magnitude\(mg\).
    2. \(\what N\), the upward normal force exerted by the spring balance, which is the weight measured by the balance.
    3. \(\what F_I\), an inertial force of magnitude\(breasts\)which is exerted upwards (in the opposite direction to the acceleration of the reference frame).

    The forces in the elevator frame are illustrated in the figure.Figure 5.6.4, along with a coordinate system that was chosen so that the forces are collinear with one of the axes (since the acceleration is zero).

    5.6: Non-Inertial Frame of Reference and Inertial Forces (5)

    All the forces are in the vertical direction, so we just need to write the\(y\)component of Newton's Second Law, which we can easily solve for the normal force: \[\begin{aligned} \sum F_y = N+F_I-F_g &=0\\ N + ma -mg &=0\\ \so N =m(g-a)\end{aligned}\] Remember to be careful with the signs. We include the fact that\(F_I\)is exerted upwards with the plus sign in the first equation (the\(y\)component of\(\vec F_I=0\chat x+F_I\hat y\)es\(+F_I\)). So we use the fact that the magnitude of the inertial force is given by\(F_I=en\)on the second line.

    You can easily verify that you would get the same result in the ground inertial frame, where there is no inertial force but the acceleration is non-zero (and in the negative\(y\)direction if we use the same coordinate system): \[\begin{aligned} \sum F_y =N-mg = -ma \quad\quad\text{(Earth frame of reference)}\end{aligned}\] O normal strength, which corresponds to the weight read on the scale, is like this\(N=m(g-a)\). We must ask ourselves if the result makes sense:

    • From the dimension of\(a\)y\(grams\)They are the same\(m(g-a)\)it has the right amount of strength.
    • if the acceleration\(a\), is zero, so the magnitude is\(N=mg\), as it would be if the elevator were at rest relative to the ground.
    • if the acceleration\(a\)It's the same as\(grams\), the normal force exerted by the scale is exactly zero and your measured weight is zero. This is what we call “weightless”, which is not a good description, as the force of the weight is still applied, and it is the normal force that is zero.
    • if the acceleration\(a\), is bigger than\(grams\), then the normal force would be negative. This corresponds to the elevator accelerating downwards faster than gravity, and the model breaks, since in this case you would first hit the ceiling of the elevator, which would exert a downward normal force of magnitude\(m(a+g)\).
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