A force , F, applied to a 16.7 kg crate at an angle, theta, of 31.0 degrees make
ID: 2126028 • Letter: A
Question
A force, F, applied to a 16.7 kg crate at an angle, theta, of 31.0 degrees makes the crate move horizontally with a constant acceleration of 1.59 m/s^2. The coefficient of kinetic friction between the crate and the surface is mu_k=0.37. Calculate the magnitude of F.
A force, F, applied to a 16.7 kg crate at an angle, theta, of 31.0 degrees makes the crate move horizontally with a constant acceleration of 1.59 m/s^2. The coefficient of kinetic friction between the crate and the surface is mu_k=0.37. Calculate the magnitude of F.Explanation / Answer
Net down force on the crate = 16.7*9.8 - F*Sin(31 degree) N
Net frictional force on crate = mu_k * net down force = 0.37 (16.7*9.8 - F*Sin(31 degree)) N
Horizontal pulling force on crate = F*cos(31 degree)
Horizontal pulling force on crate - Net frictional force on crate = crate mass* acceleration
F*cos(31 degree) - 0.37 (16.7*9.8 - F*Sin(31 degree)) = 16.7 * 1.59
F[cos(31 degree) + 0.37*sin(31 degree)] = 60.5542
F = 60.5542 / 0.8571 = 70.840 N
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