# Statics - Hanging mobile

Problem Statement:

A mobile consists of a homogeneous rod of mass m = 0.02 kg and length l = 0.4 m attached to the ceiling by a rope of negligible mass. Two objects (considered punctual) are suspended as indicated in the figure. The mass of object 1 is m1 = 0.06 kg. If the distance x shown in the figure is 0.1 m, determine the value of the mass m2 so that the system is in static equilibrium at the indicated position. Calculate the tension of the rope.

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Solution:

To solve this problem we will first draw the external forces that act on the system represented in the figure: the bar + m1 + m2. Each of these three objects undergoes the action of the gravitational force of the Earth (weight). There is also the tension of the rope that attaches the mobile to the ceiling. The tension that acts on the rope between m2 and the rod of the mobile is not taken into account because it is an internal force (the force and its reaction act on the system).

To ease the representation of these forces we will consider that the masses are punctual. The weight of the rod is applied at its center of mass. As it is homogeneous, the center of mass of the rod is at a distance l/2 from the end of the rod. We have represented the external forces acting on the system in the figure below:

For a system to be in static equilibrium, two conditions must be met:

The first condition (which is Newton’s second law applied to a system) implies that the center of mass of the system has no acceleration (linear), so the system does not move.

The second condition implies that the system does not have angular acceleration, so the system does not rotate. This condition must be met regardless of the point we choose to calculate the torques (or moments) of external forces.

In the previous figure we have represented the external forces that act on the system. Therefore, for the resultant of these forces to be zero, the following must be fulfilled:

We are now projecting the previous equation on the y axis. It is not necessary to project on the other two axes because all the forces are vertical in this problem. The positive direction of Cartesian axes is indicated on the left side of the figure.

Note that in equation (1) the signs of the projections have been taken into account, and the fact that the projection of a vector on an axis is a scalar magnitude.

Equation (1) has two unknowns variable (T and m2). Therefore we need an additional equation to solve the problem. We will obtain this equation by applying the second condition of a static equilibrium.

The previous condition must be fulfilled for any point that we choose to calculate the torque of the forces. In this case we will use the point O of the figure:

The torque (or moment) of any force F is given by:

Where r is a vector that goes from the point we have chosen as the origin to calculate the torque to the point of application of force F. Like all cross products, the previous vector has a direction given by the right-hand rule and its magnitude is:

Where θ is the angle formed by the two vectors.

We will now determine the torque of each of the forces involved in the problem.

Tension T: The tension is applied at point O, so r = 0 for this force and its torque is null.

Weight P1: The torque of P1 is:

We have represented both vectors in the following figure:

Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and inward (-k).

The torque magnitude is given by:

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Weight P2: The torque of P2 is:

We have represented both vectors in the following figure:

Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and outward (k).

The torque magnitude is given by:

As you can see in the previous figure, to calculate the torque of P2 we have not taken into account the rope from which the mass is hung m2. The following figure explains why.

If we use the rT vector to calculate the P2 moment, we can write it as the sum of vectors r and rp. But in this cross product, the term that corresponds to rp vanishes because rp is parallel to P2; and the cross product of two parallel vectors is null:

Weight P: The torque of P is:

We have represented both vectors in the following figure:

Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and inward (-k).

The torque magnitude is given by:

In the following figure we have represented the torques of the three weights (they are not scaled) in three dimensions :

We now have all the information we need to use the torques equation:

Since they are all on the z axis, we are going to project the previous equation on that axis:

And after replacing with the magnitude of each of the torques we get:

From equation (2) and substituting the givens we obtain the value of the mass m2:

From equation (1) we obtain the tension of the rope:

In the problem we have used g = 10 m/s2.

Do not forget to include the units in the results of the problems.

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