Questions: A uniform 33.0-kg beam of length l=5.40 m is supported by a vertical rope located d=1.20 m from its left end as in the figure below. The right end of the beam is supported by a vertical column. (i) (a) Find the tension in the rope. N (b) Find the force that the column exerts on the right end of the beam. (Enter the magnitude.) N

Solution Steps

First, draw a free-body diagram of the beam. The forces acting on the beam are:

• Weight (mg): Acting downwards at the center of the beam (ℓ/2).
• Tension (T): Acting upwards at distance d from the left end.
• Force from the column (Fc): Acting upwards at the right end of the beam (ℓ).

We'll choose the right end of the beam as the pivot point. This eliminates the force from the column from the torque equation, simplifying the calculations.

The beam is in rotational equilibrium, so the sum of the torques about the pivot point is zero. Torque is calculated as the force multiplied by the perpendicular distance from the pivot point to the line of action of the force. Clockwise torques are negative, and counterclockwise torques are positive.

• Torque due to weight: −mg * (ℓ/2)
• Torque due to tension: T * (ℓ − d)
• Torque due to the column: 0 (since the force is applied at the pivot point)

Summing the torques and setting the sum equal to zero:

T * (ℓ − d) − mg * (ℓ/2) = 0

Rearrange the torque equation to solve for T:

T = (mg * (ℓ/2)) / (ℓ − d)

Plug in the given values: m = 33.0 kg, g = 9.8 m/s², ℓ = 5.40 m, and d = 1.20 m.

T = (33.0 kg * 9.8 m/s² * 2.70 m) / (4.20 m) T = 207.9 N

\\(\boxed{T = 207.9 \text{ N}}\\)

The beam is also in translational equilibrium, meaning the sum of the forces in the vertical direction is zero.

• Upward forces: T + Fc
• Downward forces: mg

So, T + Fc - mg = 0

Rearrange the equation to solve for Fc:

Fc = mg - T

Plug in the values: m = 33.0 kg, g = 9.8 m/s², and T = 207.9 N

Fc = (33.0 kg * 9.8 m/s²) - 207.9 N Fc = 323.4 N - 207.9 N Fc = 115.5 N

\\(\boxed{Fc = 115.5 \text{ N}}\\)

Final Answer