# There is a sign hanging off of a building. The sign itself is hung at the far end of a rod. The rod is held up by both a pivot on the building's wall, and a vertical cable attached 1 m away from the building to the rod. The cable and the rod make a right angle.

The rod has mass 30 kg, length 4 meters, the sign has mass 30 kg.

This may be confusing, so here's my feeble attempt at a picture:

img31*picoodle*com/img/img31/9/8/15/f_phypicm_bfa429c.jpg

(replace *'s with .'s)

Now, how would I find the force the pivot exerts on the rod? I know there should be no horizontal force, but I don't know how to find the vertical force.

I apologize for the repost, but I see no other way to draw attention to my unanswered question.

## To find the force that the pivot exerts on the rod, you need to consider the forces acting on the rod in equilibrium. In this case, there are two forces acting on the rod: the weight of the rod and the tension force from the cable.

First, let's analyze the weight of the rod. The weight is given by the equation:

Weight = mass * acceleration due to gravity

The mass of the rod is given as 30 kg. The acceleration due to gravity can be assumed to be approximately 9.8 m/s^2. Therefore, the weight of the rod is:

Weight = 30 kg * 9.8 m/s^2 = 294 N

Next, let's consider the tension force from the cable. Since the cable and the rod make a right angle, the vertical component of the tension force balances out the weight of the rod. This means that the vertical force exerted by the pivot should equal the vertical component of the tension force.

To find the vertical component of the tension force, we can use trigonometry. The right triangle formed by the cable, rod, and the vertical component of the tension force has a hypotenuse of 4 meters and a side adjacent to the angle of 90 degrees of 1 meter. The vertical component can be found using the equation:

Vertical component = hypotenuse * sin(angle)

In this case, the angle is 90 degrees, so sin(90) = 1. Therefore, the vertical component equals the hypotenuse, which is 4 meters.

Finally, since the vertical component of the tension force balances out the weight of the rod, the force that the pivot exerts on the rod is also equal to 4 meters:

Force by pivot = 4 meters = 4 N

Therefore, the force that the pivot exerts on the rod is 4 N.

## To find the force that the pivot exerts on the rod, you can use the principle of torque equilibrium.

First, let's define the forces acting on the rod and sign system.

1. The weight of the rod acts downwards at its center of mass.
2. The weight of the sign also acts downwards at its center of mass.
3. The force exerted by the pivot on the rod acts upwards and is perpendicular to the rod.
4. The tension in the cable acts upwards and at a 90-degree angle to the rod.

Next, let's consider torque equilibrium. Torque is the product of the force and the perpendicular distance from the axis of rotation. Since the rod and the sign are in rotational equilibrium, the sum of their torques about any point must be zero.

Taking moments about the pivot point, we can balance the torques.

1. The weight of the rod produces a clockwise torque. The torque can be calculated as T_rod = (mass of rod) x (distance to the pivot point).
T_rod = 30 kg x 9.8 m/s^2 x (4 m / 2) = 588 Nm

2. The weight of the sign produces a clockwise torque as well. The torque is given by T_sign = (mass of sign) x (distance to the pivot point).
T_sign = 30 kg x 9.8 m/s^2 x 4 m = 1176 Nm

3. The tension in the cable produces an anti-clockwise torque. The torque can be calculated as T_cable = (tension in cable) x (distance to the pivot point). Since the cable is perpendicular to the rod, the distance is 1 m.
T_cable = Tension x 1 m

Using the principle of equilibrium, the sum of the torques should be zero, so:

T_rod + T_sign - T_cable = 0

Solving for the tension in the cable, we get:

T_cable = T_rod + T_sign
T_cable = 588 Nm + 1176 Nm
T_cable = 1764 Nm

So, the tension in the cable is 1764 Newton-meters (Nm). The force exerted by the pivot on the rod is the same as the tension in the cable, which is 1764 Newtons (N).