A uniform bar of length 12 m and mass 3 kg is attached to a wall with a hinge that exerts a horizontal force Hx and a vertiscal force Hy on the bar. The bar is held by a cord that makes a 90 degree angle with respect to bar and 43 degrees with respect to wall. What is the magnitude of the horizontal force Hx on the pivot? The acceleration of gravity is 9.8 m/s^2. Answer in units of N.
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To find the magnitude of the horizontal force Hx on the pivot, we can first break down the forces acting on the bar.
Given:
Length of the bar (L) = 12 m
Mass of the bar (m) = 3 kg
Angle between the cord and the bar (θ1) = 90 degrees
Angle between the cord and the wall (θ2) = 43 degrees
Acceleration due to gravity (g) = 9.8 m/s^2.
We can start by resolving the forces vertically and horizontally.
1. Resolving vertically:
The vertical force components can be divided into two parts:
- The weight of the bar acting downwards (mg)
- The vertical force exerted by the hinge (Hy) pointing upwards
Since there is no vertical acceleration in this case (the bar doesn't move vertically), these two forces balance each other out:
Hy = mg.
2. Resolving horizontally:
Now, let's resolve the horizontal forces.
Since the cord makes a 90-degree angle with the bar, it provides a force that holds the bar from falling.
This force can be divided into two components:
- Horizontal force exerted by the cord (Hx)
- Horizontal component of the vertical force exerted by the hinge (Hy * cos(θ2))
Since there is no horizontal acceleration in this case (the bar doesn't move horizontally), these two forces balance each other out:
Hx = Hy * cos(θ2).
Now, let's calculate Hy:
Hy = mg.
Substituting this value back into the equation for Hx:
Hx = Hy * cos(θ2)
= mg * cos(θ2).
Finally, let's calculate the magnitude of Hx by substituting the given values into the equation:
Hx = (3 kg) * (9.8 m/s^2) * cos(43 degrees).
Using a calculator, the magnitude of Hx is approximately 18.66 N.