Two blocks connected by a massless string slide down an inclined plane having an angle of inclination of 37 ° . The masses of the two blocks are M 1 = 4     k g and M 2 = 2     k g respectively and the coefficients of friction of M 1 and M 2 with the inclined plane are 0.75 and 0.25 respectively. Assuming the string to the taut, find the tension in the string. ( sin 37 ° = 0.6 , &#82

Step-by-step solution

For mass M 1 = 4     k g The component of weight acting along the incline = M 1 g     sin 37 ° = 4 × 9.8 × 0.6 = 23.52     N The Maximum frictional force = μ M 1 g cos 37 ° = 0.75 × 4 × 9.8 × 0.8 = 23.52     N Since the maximum frictional force is equal to the component of weight, so if M 1 M 2 , it won’t move and the string will become slack. For the string to remain taut the arrangement is as shown in the fig Fo