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two crates, of mass m1 = 71kg and m2 = 133kg , are in contact and at rest on a horizontal surface. Force F = 610N is exerted on the 71-kg crate. The coefficient of kinetic friction is 0.18

1.)Determine the acceleration of the system.

2.)Determine the magnitude of the force that each crate exerts on the other.

3.) If the crates are reversed, determine the acceleration of the system.

4.) If the crates are reversed, determine the magnitude of the force that each crate exerts on the other.

Added Thu, 30 Jul '15

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1. a = F net /m(net)

Fnet = 610 - umg = 610 - .18*(71+133)*9.8 = 250.144 N

so a = 250.144/(71+133) =1.226 m/s^2

2. for 133 kg block normal force -friction = 133*a

N - .18*133*9.8 = 133*1.226

N = 397.67 N

3. a = F net /m(net)

Fnet = 610 - umg = 610 - .18*(71+133)*9.8 = 250.144 N

so a = 250.144/(71+133) =1.226 m/s^2

4. for 71 kg block normal force -friction = 71*a

N - .18*71*9.8 = 71*1.226

N = 212.29 N

Added Thu, 30 Jul '15

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