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A block of mass m = 5kg is place on inclined plane of angle $\lambda = 37^o$ . A constant force P is applied of an angle $\alpha = 53^o$ as shown in diagram . What is the value of minimum force of P(in N) , so that the block does not slide down the plane. $\left ( g = 10\frac{m}{s^2} \right )$ .

Option: 1
50

Option: 2
30

Option: 3
40

Option: 4
70

Answers (1)

best_answer

As we learn

Minimum Force to avoid sliding of body down on an inclined plane -

$P=W \left[\frac{sin(\lambda-\theta)}{cos(\theta+\alpha)} \right ]$

P = Pulling force

$\lambda=$ Angle of Inclination

$\theta=$ Angle of friction

- wherein

For equilibrium

$R+P sin\alpha=W cos\lambda$

$P cos \alpha+F=W sin \lambda$

Use $F=\mu R$

$P=\frac{Wsin(\lambda-\theta)}{cos(\theta+\alpha)}$

So For minimum value of force p

$f_{L}=mg \sin \lambda - P \cos \alpha -(i)$

$R=mg\cos \lambda - P \sin \alpha$

$f_{L}=\mu R = \mu (mg\cos \lambda -P \sin \alpha )-(ii)$

By (i) and (ii)

$mg\sin \lambda -p \cos \alpha=\mu(mg\cos \lambda -p \sin \alpha)$

$p(\cos \alpha -\mu \sin \alpha)=mg\sin\lambda -\mu mg \cos \lambda$

$p=\frac{mg(\sin \lambda -\mu \cos \lambda)}{\cos \alpha -\mu sin\alpha }$

$\Rightarrow p=\frac{5*10\left [ \frac{3}{5}-\left [ 0.6*\frac{4}{5} \right ] \right ]}{\left [ \frac{3}{5}-\left [ 0.6*\frac{4}{5} \right ] \right ]}= 50 N$

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