Question

3. What are the tension of the rope and the force made by the pivot on the beam (magnitude and direction) in this situation?

Answer 1

Hi there!

We can use a summation of torques to solve.

Recall the following:

`\Sigma \tau = r * F\\\\`

For a system to be static:

`\Sigma \tau = 0 \\\\\Sigma F = 0`

In this case, the counterclockwise torques must sum up to the clockwise torques. We can use the pivot as our fulcrum.

Clockwise torques:
1. Weight of man (F = Mg = 70 · 9.8 = 686 N)

2. Weight of beam (F = Mg = 20 · 9.8 = 196 N)

Counterclockwise torque:
3. VERTICAL comp. of tension (F = Ty = Tsin(φ))

Do a summation of torques. The 'r' value is the distance from the force's line of action to the pivot.

Ex: For the beam, its center of mass (assuming uniform density) is at 1.5 m, or its center.

`\tau_(cc) = \tau_(ccw)\\\\rF_1 + rF_2 = rF_3\\\\(1)(686) + 1.5(196) = 3(Tsin(30))}}\\\\686 + 294 = 3T(0.5)\\\\980 = 1.5T\\\\T = \boxed{653.33 N}`

Now, we can use a summation of forces to determine the vertical and horizontal components of the pivot's force.

`\Sigma F_y = 0`

Sum the forces in the vertical direction. Let 'V' represent the pivot's vertical force.

`0 = T_y + V - W_m - W_b\\\\0 = Tsin(30) + V - 686 - 196\\\\882 = 326.67 + V\\\\V = 555.34 N`

Now, sum the horizontal forces. 'H' is the pivot's horizontal force.

`\Sigma F_x = 0 \\\\0 = T_x - H \\\\H = Tcos(30)\\\\H = 653.33cos(30) = 565.80 N`

These are the components, so use Pythagorean Theorem to find the total pivot force.

`F_p = √(V^2 + H^2) = \boxed{792.80 N}`

The force will point towards the UPPER LEFT (diagonal). We can solve for the exact angle:

`tan\theta = V/H \\\\tan{-1}(555.34/565.80) = \boxed{44.47^o}`