91Ó°ÊÓ

A uniform strut of mass \(m\) makes an angle \(\theta\) with the horizontal. It is supported by a frictionless pivot located at one- third its length from its lower left end and a horizontal rope at its upper right end. A cable and package of total weight \(w\) hang from its upper right end. (a) Find the vertical and horizontal components \(V\) and \(H\) of the pivot's force on the strut as well as the tension \(T\) in the rope. (b) If the maximum safe tension in the rope is 700 \(\mathrm{N}\) and the mass of the strut is \(30.0 \mathrm{kg},\) find the maximum safe weight of the cable and package when the strut makes an angle of \(55.0^{\circ}\) with the horizontal. (c) For what angle \(\theta\) can no weight be safely suspended from the right end of the strut?

Step by step solution

01

Analyze Forces on the Strut

Identify all the forces acting on the strut. These include the tension in the rope (\( T \)), the weight of the strut (\( mg \); acting downwards at the center of mass located at L/2 vertically down), the weight of the package and cable (\( w \)), and the forces at the pivot (\( H \) and \( V \)).

02

Apply Static Equilibrium Conditions

Since the strut is in static equilibrium, the sum of forces in horizontal and vertical directions, and the sum of moments about any point, must all equal zero. We will consider the pivot point to reduce variables.

03

Sum of Vertical Forces

Applying the equilibrium condition for vertical forces:\[V - mg \cos(\theta) - w = 0\]Hence, the vertical component of the pivot force is:\[ V = mg \cos(\theta) + w \]

04

Sum of Horizontal Forces

Applying the equilibrium condition for horizontal forces:\[H - T = 0\]Thus, the horizontal component of the pivot force is:\[ H = T \]

05

Sum of Moments about the Pivot

Taking moments about the pivot and assuming clockwise moments as positive, we have:\[T \cdot \frac{2}{3}L = mg \cdot \frac{1}{3}L \cdot \cos(\theta) + w \cdot L \cdot \cos(\theta)\]Solve for tension \( T \) in the rope:\[ T = \frac{mg \cdot \cos(\theta) + 3w}{2} \]

06

Maximum Safe Weight Calculation

Given the maximum tension,\( T_{max} = 700 \, \mathrm{N} \)and\( m = 30 \, \mathrm{kg} = mg = 30 \times 9.8 \, \mathrm{N}\)Using:\[ T = \frac{mg \cdot \cos(55^\circ) + 3w}{2} \]Set \( T = 700 \) and solve for \( w \):\[ 700 = \frac{30 \times 9.8 \times \cos(55^\circ) + 3w}{2} \]Now, solve this equation to find \( w_{max} \).

07

Angle for No Weight Support

No weight can be supported when \( T = 0 \). From the earlier equation, setting \( T = 0 \):\[ mg \cdot \frac{1}{3} \cdot \cos(\theta) + w \cdot \cos(\theta) = 0\]Rearrange the equation:\[ \cos(\theta) = -\frac{3w}{mg}\]In this scenario, \( w = 0 \), meaning any weight results in instability, solving provides\( \theta \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Equilibrium

Understanding static equilibrium is crucial when analyzing forces that act on an object at rest. In this state, an object remains motionless and all net forces and torques are balanced. For the strut problem, the equilibrium conditions mean that:

• The sum of all horizontal forces acting on the strut equals zero.
• The sum of all vertical forces acting on the strut equals zero.
• The sum of all moments (torques) about any point is zero.

In essence, if an object is in static equilibrium, none of its parts are accelerating. This concept allows us to solve for unknown forces acting on the strut in different directions.

Newton's Laws

Newton's Laws, particularly the first and third laws, are fundamental in force analysis. Newton's First Law states that an object remains at rest or in uniform motion unless acted upon by an external force. This applies directly to static situations where objects like our strut don't accelerate or move because the forces perfectly balance out.

Newton's Third Law, often phrased as "for every action, there is an equal and opposite reaction," helps us understand interactions, such as the force of the rope on the strut being equal in magnitude but opposite to the strut's pulling force. These laws serve as the legal framework when calculating forces in static equilibrium scenarios.

Force Analysis

In force analysis, we identify and analyze all forces acting on the object. For the strut problem:

• The weight of the strut itself acts downward through its center of gravity.
• The tension in the rope acts horizontally at the upper right end.
• There is a vertical and horizontal force at the pivot point.
• The weight of the cable and package acts downward at the upper end.

By setting up equations from the equilibrium conditions, we can solve for unknown forces, such as the tension in the rope or the force components at the pivot. This involves carefully balancing and breaking down forces into horizontal and vertical components.

Moment of Forces

Moments of forces, or torques, explain how forces cause rotations about a pivot or axis. In this context, a moment is calculated by multiplying the force's magnitude by the perpendicular distance from the point of rotation. For the strut problem, equilibrium also means that:

• Sum of moments about the pivot point is zero.
• This involves balancing the moments caused by the tension, the weight of the strut, and the weight of the package.

Using these moments, we can solve for unknown forces. For example, by taking moments about the pivot point, we have a direct equation involving the tension in the rope and can calculate it accordingly. Understanding moments is essential because it allows us to see how forces influence rotational equilibrium.