A uniform 8 m 1500 kg beam is hinged to a wall and supported by a thin cable attached 2 m from the free end of the beam as shown in the figure. The beam is supported at an angle of 30 degrees above the horizontal.

a) free diagram
b) find the tension in the cable
c) how hard does the beam push inward on the wall?

a) I'm sorry, I don't have the capability to show you a figure. However, I can try to describe it to you. Imagine a beam that is 8 meters long, with one end hinged to a wall. The other end is supported by a cable that is attached 2 meters from the free end of the beam. The beam is angled 30 degrees above the horizontal.

b) To find the tension in the cable, we need to consider the forces acting on the beam. The weight of the beam can be split into two components: one parallel to the cable, and one perpendicular to it. The component parallel to the cable will contribute to the tension, while the component perpendicular to the cable will contribute to the force pushing inward on the wall.

To find the tension in the cable, we can use the equation:

Tension = Weight × tangent(angle)

The weight of the beam can be calculated using its mass and the acceleration due to gravity:

Weight = mass × gravity

Combining these equations, we can find the tension in the cable.

c) The beam pushes inward on the wall with a force equal to the component of its weight perpendicular to the cable. This force can be calculated using the equation:

Force = Weight × sine(angle)

Again, we can use the weight of the beam calculated in part b to find the force pushing inward on the wall.

a) To draw a free diagram of the beam, we need to represent all the forces acting on it.

1) The weight of the beam, which acts vertically downward at its center of mass.
2) The tension in the cable, which acts upward at an angle of 30 degrees above the horizontal.
3) The normal force exerted by the wall, which acts perpendicular to the wall.

Here is the free diagram of the beam:

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b) To find the tension in the cable, we can use the principle of equilibrium. Since the beam is not rotating, the sum of the horizontal and vertical forces acting on it must be zero.

Sum of vertical forces:
Tension - Weight of the beam * cos(30) = 0

Sum of horizontal forces:
Weight of the beam * sin(30) - Normal force = 0

From the first equation, we can solve for the tension:
Tension = Weight of the beam * cos(30)

The weight of the beam can be calculated as the product of its mass and the acceleration due to gravity:
Weight of the beam = mass * gravity

Given:
mass of the beam = 1500 kg
gravity = 9.8 m/s^2

Weight of the beam = 1500 kg * 9.8 m/s^2

Now we can calculate the tension:

Tension = (1500 kg * 9.8 m/s^2) * cos(30)

c) How hard the beam pushes inward on the wall can be determined by the normal force exerted by the wall. From the second equation above, we can solve for the normal force:

Normal force = Weight of the beam * sin(30)

Normal force = (1500 kg * 9.8 m/s^2) * sin(30)

a) To draw a free diagram, we need to represent the given information visually. The beam is hinged to a wall and supported by a cable at an angle of 30 degrees above the horizontal. We can represent the beam as a straight line in the diagram. The wall should be represented as a vertical line. We also need to draw the cable, which connects the beam to the wall at a distance of 2 m from the free end of the beam.

b) To find the tension in the cable, we can use the concept of equilibrium. Since the beam is not moving, the sum of the forces acting on it in both the vertical and horizontal directions should be zero. Let's denote the tension in the cable as T.

In the vertical direction, the tension in the cable (T) and the weight of the beam (W) act in opposite directions. The weight of the beam can be calculated using the formula W = mg, where m is the mass of the beam and g is the acceleration due to gravity (approximately 9.8 m/s^2). Since the beam is uniform, its weight is evenly distributed. Therefore, we can calculate the weight of the beam using the formula W = (mass of the beam / length of the beam) * length of the segment.

In the horizontal direction, there are no other forces acting on the beam. Therefore, the tension in the cable is the only force in this direction.

Using these equations, we can solve for T.

c) To find how hard the beam pushes inward on the wall, we need to consider the horizontal component of the force exerted by the beam on the wall. This force acts perpendicular to the wall. From the free diagram, we can see that the horizontal component is equal to the tension in the cable.

Now, we can use the equations derived in part b) to find the tension in the cable, and then calculate the force exerted by the beam on the wall.

Tension= 13203.5

beam pushes at magnitude 13002.91N
My sig figs maybe be slightly off

Without the diagram, it is not clear at what angle the cable makes with the vertical.

Draw the free diagram.

The hinge at the wall means that there will be a horizontal and a vertical force to achieve equilibrium, meaning
ΣFx=0
ΣFy=0, and
Σmoments=0

The "uniform beam" means that the weight is distributed uniformly along the beam, or its centre of mass is at mid-length.

You will have three unknowns, namely Rx and Ry at the hinge, and T for the tension of the cable.

Set-up the 3 equilibrium equations and solve for the unknowns, and hence the solution to (b) and (c).