A uniform beam of length L = 2.2 m and mass M = 49 kg has its lower end fixed to pivot at a point P on the floor, making an angle θ = 18° as shown in the diagram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam one-fourth L from its upper end.

What is the y-component Py of the force, in newtons, exerted by the pivot on the beam?

Write an expression for the tension T in the horizontal cable AB.

What is the x-component Px of the force, in newtons, exerted by the pivot on the beam?

I cannot figure part C

We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)

These sum to zero because the beam is not accelerating vertically.

[tex]\Sigma F = -F_g - T + P_y = 0[/tex]

[tex]P_y = F_g + T[/tex]

The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.

[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]

Part B.

We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)

- The tension of the horizontal cable acting at the end of the rod (CCW, at L)

- The force of tension in the rope holding the box (CC, at 3L/4)

Since the rod is not rotating, these torques sum to zero.

The equation for torque is:
[tex]\tau = r \times F[/tex]

This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.

Now, let's find the torque from all three of these forces.

- Force of gravity at center:

The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.

[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]

- Tension of horizontal cable:

The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.

[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]

- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]

Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]

Part C.

This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.

We only have two:
- The horizontal tension in the cable to the left (1847.38 N)

- The horizontal force exerted by the pivot on the beam to the right

These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]

**to the right