Math Problem Statement

A uniform 34.5-kg beam of length ℓ = 4.10 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. A beam of length ℓ is suspended from a rope and supported by a vertical column. The rope is attached at a point that is a distance d from the left end of the beam. The column is beneath the right end of the beam. (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

To solve this problem, we need to analyze the forces and torques acting on the beam to find the tension in the rope and the force exerted by the column.

Given:

  • Mass of the beam, m=34.5m = 34.5 kg
  • Length of the beam, =4.10\ell = 4.10 m
  • Distance from the left end of the beam to the point where the rope is attached, d=1.20d = 1.20 m
  • Gravitational acceleration, g=9.81g = 9.81 m/s²

(a) Finding the tension in the rope:

  1. Identify forces:

    • Weight of the beam, W=mg=34.5×9.81=338.145W = mg = 34.5 \times 9.81 = 338.145 N, acting at the center of the beam (at /2=2.05\ell/2 = 2.05 m from the left end).
    • Tension in the rope, TT, acting upward at d=1.20d = 1.20 m from the left end.
    • Force exerted by the column, FcF_c, acting upward at the right end (at =4.10\ell = 4.10 m from the left end).
  2. Set up the torque equilibrium about the point where the column supports the beam (right end): Sum of torques about the right end=0\text{Sum of torques about the right end} = 0 The torque due to the weight: τweight=W×(2.054.10)=338.145×(2.05)=693.20225Nm\tau_{\text{weight}} = W \times (2.05 - 4.10) = 338.145 \times (-2.05) = -693.20225 \, \text{N} \cdot \text{m} The torque due to the tension: τtension=T×(1.204.10)=T×(2.90)Nm\tau_{\text{tension}} = T \times (1.20 - 4.10) = T \times (-2.90) \, \text{N} \cdot \text{m}

    Setting up the equation for equilibrium: τtension+τweight=0\tau_{\text{tension}} + \tau_{\text{weight}} = 0 T×(2.90)693.20225=0T \times (-2.90) - 693.20225 = 0 T=693.202252.90=239.04NT = \frac{693.20225}{2.90} = 239.04 \, \text{N}

So, the tension in the rope is approximately T=239.04T = 239.04 N.

(b) Finding the force that the column exerts on the right end of the beam:

  1. Force equilibrium in the vertical direction: Sum of vertical forces=0\text{Sum of vertical forces} = 0 The forces acting vertically are:

    • Upward force by the rope, T=239.04T = 239.04 N
    • Upward force by the column, FcF_c
    • Downward weight of the beam, W=338.145W = 338.145 N

    Setting up the equation: T+Fc=WT + F_c = W 239.04+Fc=338.145239.04 + F_c = 338.145 Fc=338.145239.04=99.105NF_c = 338.145 - 239.04 = 99.105 \, \text{N}

So, the force that the column exerts on the right end of the beam is approximately Fc=99.105F_c = 99.105 N.

Summary:

  • (a) Tension in the rope, T=239.04T = 239.04 N
  • (b) Force by the column, Fc=99.105F_c = 99.105 N

Would you like a detailed explanation on any of these steps or have any questions? Here are some related questions you might find interesting:

  1. How would the tension change if the rope was moved further to the right?
  2. How would the forces change if the mass of the beam increased?
  3. What if the beam was not uniform; how would you calculate the torques?
  4. How would this problem change if there was an additional force acting horizontally?
  5. What is the significance of choosing the right end as the pivot point?

Tip: When analyzing forces and torques, always pick a pivot point where unknown forces can be eliminated from the torque equation to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Statics
Torque
Equilibrium of forces

Formulas

Torque equation: τ = F × d
Force equilibrium: ΣF = 0
Torque equilibrium: Στ = 0
Gravitational force: W = mg

Theorems

Principle of static equilibrium

Suitable Grade Level

Undergraduate Physics