The coefficient of static friction between the m = 3.50 kg crate and the 35.0° incline of the figure below is 0.270... ThanksWhat minimum...

Given:

mass = m = 3.5 kg

Coefficient of friction = F = 0.270

Angle of incline = A = 35.0 degrees.

Assumed:

Acceleration due to gravity = g = 9.81 m/s^2

Solution:

The force perpendicular to the incline due to the component of weight of crate =

f1 = m*g*Cos A = 3.5*9.81*Cos 35 = 3.5*9.81*0.8192 = 28.1272 N

The force parallel to the incline due to the component of weight of crate =

f2 = m*g*Sin A = 3.5*9.81*Sign 35 = 3.5*9.81*0.5736 = 19.694556 N

This force (f2) is pulling the crate down along the incline.

The frictional force preventing this movement =

f3 = f1*F = 28.1272*0.27 = 7.59435264 N

Additional frictional force required prevent the movement = f4 = f2 - f3 = 19.694556 - 7.59435264 = 12.10020336 N

Additional force vertical to incline required to provide this additional frictional force f4 = f5 = f4/0.270 = 44.815568 N

Answer:

Minimum force required = 44.815568 N