A cart of mass mA = 5.9 kg is pushed forward by a horizontal force F. A block of

Question

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 ?oor without falling down. Given g = 9.8 m/s^ 2
and the static friction
coe?cient 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 in units of N

Explanation / Answer

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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