1.2 kN H = 0.35 Hk = 0.25
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
Solve Prob. 4.75 when θ = 30° and P = 150 N.
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,
Coefficient of frictions,
Static coefficient ()=0.35
Kinematic coefficient ()= 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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