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1.2 kN H = 0.35 Hk = 0.25

1.2 kN H = 0.35 Hk = 0.25

Elements Of Electromagnetics
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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Solve Prob. 4.75 when θ = 30° and P = 150 N.

1.2 kN H = 0.35 Hk = 0.25
Transcribed Image Text:1.2 kN H = 0.35 Hk = 0.25
Expert Solution
Step 1

Given data:

The block resting on an inclined plane

The force F, acting on the block = 1.2 kN = 1200 N( vertically downwards)

The load P, acting on the block = 150 N, (which is acting horizontally)

Angle, θ=300

Coefficient of frictions, 

Static coefficient (μs)=0.35

Kinematic coefficient (μk)= 0.25

From the figure, it is to be observed that the forces applied on block to maintain the equilibrium.

Therefore, we need to find the magnitude and direction of the frictional force.

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