Answered: 50 1Ь Hj = 0.40 Hi = 0.30 40° 30°

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

Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when P = 120 lb.

Expert Solution

Step 1

Given:

The magnitude of force P is P= 120 lb.
The static coefficient of friction is μ_s=0.40.
The coefficient of friction is μ_k=0.30.

Here R is a normal force to balance the force due to weight of block and F is the force to balance out the force due to P.

The expression for the sum of forces in horizontal direction,

$\begin{array}{rcl} \sum F_{X} & = & 0 \\ - F + 50 lb \times \sin 30 ° - 120 lb \times \cos 40 ° & = & 0 \\ - F + 25 lb - 91.2 lb & = & 0 \\ F & = & - 66.2 lb \end{array}$

The expression for the sum of forces in the vertical direction,

$\begin{array}{rcl} \sum F_{y} = 0 R - 50 lb \times \cos 30 ° - 120 lb \times \sin 40 ° = 0 R - 43.3 lb - 77.13 lb = 0 R = 120.13 lb \end{array}$

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