Answered

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Two blocks are connected by a pulley system represented in the diagram below. M1= 60kg and M2=12kg. Wh
is the coefficient of friction required to keep the masses at rest?

Sagot :

leena

Hi there!

Assuming the pulley is frictionless and massless, we can do a summation of forces for each block.

Block 1 has the tension force (in direction of acceleration) and force of static friction acting on it. For the block to be stationary, the net force must be 0, so:
[tex]\Sigma F = T - F_S[/tex]

[tex]0 = T - F_S\\\\T = F_S[/tex]

Now, we can do a summation for Block 2.

[tex]\Sigma F = W_2 - T\\\\0 = W_2 - T\\\\T = W_2[/tex]

We can substitute this tension in the above equation to solve for the force of static friction and the coefficient of static friction.

[tex]F_S = W_2\\\\\mu m_1g = m_2g\\\\\mu = \frac{m_2}{m_1} = \boxed{\mu = .2}[/tex]

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