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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitude of the largest and smallest force Q for which equilibrium is maintained when 0 = 30°, I= 0.2 m, and P= 40 N.

Expert Solution

Step 1 of 2.

To find:

The largest and smallest force.

Given:

The length of the bar is $l = 0.2 m$ .

The angle is $θ = 30 °$ .

The load is $P = 40 N$ .

The coefficient of static friction is $μ = 0.15$ .

FBD:

The free body diagram is shown below.

Step 2 of 2.

Calculation:

Take a moment about the point $B$ .

$\begin{array}{rcl} \sum_{} M_{B} & = & 0 \\ - A x_{A} + \frac{x_{A}}{2} P & = & 0 \\ A & = & \frac{40}{2} \\ = & 20 N \end{array}$

The friction force can be obtained as

$\begin{array}{rcl} F & = & μ A \\ = & 0.15 \times 20 \\ = & 3 N \end{array}$

and

$\begin{array}{rcl} x_{A} & = & 2 l \sin θ \\ δ x_{A} & = & 2 l \cos θ δ θ \\ y_{F} & = & 3 l \cos θ \\ δ y_{F} & = & - 3 l \sin θ δ θ \end{array}$

The largest force can be obtained by virtual work as

$\begin{array}{rcl} δ U & = & 0 \\ (Q_{m a x} - F) δ x_{A} + P δ y_{F} & = & 0 \\ (Q_{m a x} - 3) \times (2 \times 0.2 \times \cos 30 \times δ θ) + 40 \times (- 3 \times 0.2 \times \sin 30 \times δ θ) & = & 0 \\ Q_{m a x} & = & 37.64 N \end{array}$

The smallest force can be obtained as

$\begin{array}{rcl} (Q_{m i n} + F) δ x_{A} + P δ y_{F} & = & 0 \\ (Q_{m i n} + 3) \times (2 \times 0.2 \times \cos 30 \times δ θ) + 40 \times (- 3 \times 0.2 \times \sin 30 \times δ θ) & = & 0 \\ Q_{m i n} & = & 31.64 N \end{array}$

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