A uniform 22 kg beam, 3.0 m long, is attached to a wall on the left by a hinge, and is supported by a cable on its right end. A 13 kg monkey hangs from the beam, 1.0 m from the left end.a) Determine the tension in the cable.b) Determine the magnitude of the force due to the hinge on the beam. 1. A uniform 22 kg beam, 3.0 m long, isattached to a wall on the left by a hinge, andis supported by a cable on its right end. A 13kg monkey hangs from the beam, 1.0 m fromthe left end.25°7a) Determine the tension in the cable.b) Determine the magnitude of the force dueto the hinge on the beam.Fr

Answered: 1. A uniform 22 kg beam, 3.0 m long, is…

Physics for Scientists and Engineers: Foundations and Connections

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Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter14: Static Equilibrium, Elasticity, And Fracture

Section: Chapter Questions

Problem 33PQ: Problems 33 and 34 are paired. One end of a uniform beam that weighs 2.80 102 N is attached to a...

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A uniform 22 kg beam, 3.0 m long, is attached to a wall on the left by a hinge, and is supported by a cable on its right end. A 13 kg monkey hangs from the beam, 1.0 m from the left end.

a) Determine the tension in the cable.

b) Determine the magnitude of the force due to the hinge on the beam.

1. A uniform 22 kg beam, 3.0 m long, is
attached to a wall on the left by a hinge, and
is supported by a cable on its right end. A 13
kg monkey hangs from the beam, 1.0 m from
the left end.
25°7
a) Determine the tension in the cable.
b) Determine the magnitude of the force due
to the hinge on the beam.
Fr

Expert Solution

Step 1 :Introduction

Torque acting on an object is defined as the product of perpendicular force to the distance from the pivot point , that is , $τ = r F$ .

Let the hinge point be the pivot point , so the torques are due to the tension force acting in the anti-clockwise direction , which is , $τ$ acting at a distance $d_{τ} = 3 m$ from the pivot point , the weight force of the beam acting clockwise direction , which is , $F_{b} = m_{b} g$ acting at a distance $d_{b} = 1.5 m$ from the pivot point and the weight force of the monkey acting clockwise direction , which is , $F_{m} = m_{m} g$ acting at a distance $d_{m} = 1 m$ from the pivot point . As the system is at equilibrium , the net torque will be zero , that is , $\begin{array}{rcl} T \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} _{T} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} m \end{array} \begin{array}{rcl} _{b} \end{array} \begin{array}{rcl} g \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} _{b} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} m \end{array} \begin{array}{rcl} _{m} \end{array} \begin{array}{rcl} g \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} _{m} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 0 \end{array}$ , where $m_{b} = 22 k g$ is the mass of beam , $m_{m} = 13 k g$ is the mass of monkey and $g = 9.8 m / s^{2}$ is the acceleration due to gravity .

The horizontal forces acting are the horizontal component of hinge force which is $F_{H x} = F_{H} \cos (θ)$ and the the horizontal component of tension force , which is , $τ_{x} = τ \cos (25 °)$ . As the system is at equilibrium , the relation will be $F_{H} \cos (θ) = T \cos (25 °)$ .

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