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Brent0423

  • 4 years ago

A cart of mass mA = 5.9 kg is pushed forward by a horizontal force F. A block of mass mB = 0.9 kg is in turn pushed forward by the cart. If the cart and the block accelerate forward fast enough, the friction force between the block and the cart would keep the block suspended above the floor without falling down. Given g = 9.8 m/s^ 2 and the static friction coefficient µs = 0.67 between the block and the cart; the ?oor is horizontal and there is no friction between the cart and the ?oor. Calculate the minimal force F on the cart that would keep the block from falling down. Answer i

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  1. Brent0423
    • 4 years ago
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    Answer in Newtons

  2. Brent0423
    • 4 years ago
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    please help!

  3. dave444
    • 4 years ago
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    Be patient. I'm working on it.

  4. Brent0423
    • 4 years ago
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    thanks :)

  5. dave444
    • 4 years ago
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    Okay. In order to prevent slipping, the force exerted by the cart on the block must be such that the frictional force balances that of gravity. The frictional force is mu times that force. Now the force of gravity is g * m_b. So the force that must be exerted by the cart is F_c = m_b * g / mu. Now, with a free body diagram, we can see that the force exerted on the block is F_c = m_b * a, where a is the acceleration. So, a = F_c / m_b = g / mu. But if F is the force applied to the cart, F = (m_a+m_b)*a. So, F = (m_a + m_b) * g / mu. Plugging in m_a = 5.9kg, m_b = 0.9 kg, mu = 0.67, and g = 9.8 m/s^2, we get: F = 99.5 N

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