Question

A mass of 100 kg is placed on a rough inclined plane
1
of inclination angle 30°. If coefficient of friction is
13
the minimum force required parallel to plane to keep
the mass in equilibrium is (g = 9.8 m/s2)
​

Answer 1

Explanation:

dear see the image will be helpful for you...

Answer 2

Answer:

The minimum force required to plane to  keep the mass in equilibrium is 10052 N

Explanation:

Given , a mass of 100 kg is placed on a rough inclined plane

inclination angle = 30

coefficient of friction(μ) = 12  

Step 1:

Force acting down the plane

Therefore , $F_{1}$ = Mg sinx = 100 ×9.8 ×sin 30

    ⇒$F_{1}$  = 980×$\frac{1}{2}$         (value of sin30 is $\frac{1}{2}$)

  ⇒  $F_{1}$ = 490 N

Also , force of friction  = μN

$F_{2}$ = μ mg cosx

⇒ $F_{2}$ = 13 ×100 × 9.8 × $\frac{\sqrt{3} }{2}$

⇒$F_{2}$ = 12740 ×0.866 = 11032 N

Therefore ,

the minimum force  required to plane keep the mass in equilibrium =  $F_{2} - F_{1}$ = 11032 - 980 = 10052 N required .

Final answer :

Hence , 10052 N force required  to keep the mass in equilibrium .

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