**Problem Statement:**

A homogeneous rod of length L = 1 m and mass M = 0.5 kg is attached to the wall by a joint, as shown in the figure. The rod is attached by its center of mass to a rope passing through a pulley of negligible mass, from which hangs a mass m _{2} = 0.2 kg. The system is in static equilibrium. Determine:

- the tension of the rope.
- the value of the angle β.
- the components of the reaction for the joint.

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**Solution:**

First let’s draw the forces acting on the rod-mass system:

The tension of the string and the gravitational force (the weight) act on the mass *m _{2}*. The mass of the pulley is negligible, so we will not use it to solve the problem. The rod is subject to the weight, the tension of the rope and the two components of the reaction for the joint that we will calculate separately.

To calculate the tension of the string we consider only the right part of the figure, that is, the forces acting on the mass *m _{2}*. As it is at rest, Newton’s second law for this mass is:

We will now make the projection of the vectors of Newton’s second law on the *y* axis taking into account the positive direction of the axes as they are represented in the previous figure:

Since the pulley has a negligible mass, the tension of the rope that holds the rod will have the same magnitude.

To calculate the angle β and the components of the reaction for the joint we impose the two **conditions of static equilibrium** on the rod:

Taking into account the external forces represented in the previous figure that act on the rod, the first condition is:

We now project on the x and y axes taking into account the positive directions defined in the figure. We have represented the angles necessary to calculate the projections of the forces on the axes in the figure below:

The system constituted by equations (1) and (2) has three unknowns variable: β, R_{x} and R_{y}. We therefore need a third equation to solve it, which comes from imposing the second condition of static equilibrium on the rod.

This condition must be fulfilled for any point that we take as the origin of torques. To solve this problem we will use **the point A represented in the figure**.

Remember that the torque (or moment of a force) is defined as:

And its magnitude is given by:

Where θ is the angle formed by both vectors.

And where **r** is a vector that goes from the origin of moments to the point of application of the force.

The second condition of static equilibrium applied to the rod is:

The first two terms of the previous expression are null because the forces R_{x} and R_{y} are applied at point A and, therefore, the vector **r** is null for both.

We are going to calculate the two remaining torques that appear in the previous equation.

__Weight P_{1}:__ The torque of

**P**is:

_{1}We have represented both vectors in the following figure. And using the **right-hand rule**, we determine that the direction of the cross product is perpendicular to the screen and inward (**-k**).

The weight has been moved to point A in the figure (in pink) so that it is easier to see the direction of the cross product.

The torque magnitude is given by:

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__Tension T:__ The torque of

**T**is:

We have represented both vectors in the following figure. And 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:

In the following figure we have represented both torques (they are not scaled).

Therefore, we can now project the second condition of static equilibrium on the *z* axis:

And after substituting the magnitudes of the two vectors we get:

The equations (1), (2) and (3) will allow us to solve the problem. For convenience we will write the three equations together here:

From equation (3) we obtain the angle β:

From equation (1) we get R_{x}:

Finally, from equation (2) we get R_{y}:

The minus sign indicates that R_{y} has a sign opposite to the one represented in the figure.

In the problem we have used as acceleration of gravity g = 10 m/s^{2}.

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