Question #13390

a worker pushing a wheelbarrow of mass 50kg with a constant velocity along a horizontal surface whose coefficient of friction is 0.5..what is the force P exerted by the worker if angle is 30 degrees?

Expert's answer

Ff = F * k = (m * g - P * sinA) * k

Ff = P * cosA

m * g * k - P * sinA * k

= P * cosA

m * g * k = P * (cosA + sinA * k)

P = m * g * k / (cosA + sinA

* k)

P = 50 * 9.81 * 0.5 / (0.8660 + 0.5 * 0.5) = 245.25 / 1.116 = 220 N

Ff = P * cosA

m * g * k - P * sinA * k

= P * cosA

m * g * k = P * (cosA + sinA * k)

P = m * g * k / (cosA + sinA

* k)

P = 50 * 9.81 * 0.5 / (0.8660 + 0.5 * 0.5) = 245.25 / 1.116 = 220 N

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