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  • How to solve this problem- Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should

Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is :

  • Option 1)

    10.3 kg
     

  • Option 2)

    18.3 kg
     

  • Option 3)

     27.3 kg
     

  • Option 4)

    43.3 kg

 

Answers (1)

best_answer

In equilibrium

m_{1}g=\mu \left ( m+m_{2} \right )

or m=\frac{m_{1}}{\mu }-m_{2}=\frac{5}{15}-10

m=23.33kg

 

Limiting Friction -

Magnitude of limiting friction between any two bodies in contact is directly proportional to the normal reaction between them.

F_{l}\;\alpha\ R  or   f_{l}=\mu_{s}R

f_{l}= limiting friction 

\mu_{s}= coefficient of friction

R = reaction force

- wherein

*    Maximum value of static friction is limiting friction.

*     Direction is always opposite to relative motion.

 

 


Option 1)

10.3 kg
 

This is incorrect

Option 2)

18.3 kg
 

This is incorrect

Option 3)

 27.3 kg
 

This is correct

Option 4)

43.3 kg

This is incorrect

Posted by

prateek

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