Question

An effort of 180N is required just to move a certain body up an inclined plane of angle 15°,the force being parallel to the plane .if the angle of inclination to the plane is made 20°,the effort required again applied parallel th the plane is found to be 210N.find the weight of the body and coefficient of friction

Answer 1

Answer:

case I :

• F = 180 N
• friction = uN

or, f = umgcosx

• angle = 15

opposing forces , (at equilibrium)

• friction
• mg sinx

at equilibrium;

F = sum of opposing forces (friction + mg sin15)

or, F = u.mg.cos 15 + mg sin15

or, F = mg( ucos15 + sin 15)

mg = Weight (W)

Therefore , F = W( u x 0.96 + 0.26)

or,

$\sf{180 = \dfrac{w}{10}(9.6u + 2.6)}......equ(1)$

$\sf{simlarly \: for \: second \: case \: . \: 210 = \dfrac{w}{10}(9.3u + 3.4) }......equ(2)$

divide equ 2/1

$\sf{ \dfrac{7}{6} = \dfrac{9.3u \: + 3.4}{9.6u \: + 2.6} } \\ \implies \: \sf{\bold{u \: = 0.2}}$

By putting the value of u in any equation , you get w = 399 kg .

Also refer :

https://brainly.in/question/16382150