Physics!!!?
A clerk moves a box of cans down an aisle by pulling on a strap attached to the box. The clerk pulls with a force of 172.0 N at an angle of 25° with the horizontal. The box has a mass of 40 kg, and the coefficient of kinetic friction between the box and the floor is 0.45. Find the acceleration of the box.
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FBD of the box
Sum of the horizontal forces = m*a = T*cos(25) - u*N
(u*N is frictional force)
Sum of the vertical forces = 0 = T*sin(25) + N - m*g
Solve for N
N = m*g - T*sin(25)
Use that back in the horizontal equation
m*a = T*cos(25) - u*N
a = (T*cos(25) - u*(m*g - T*sin(25)))/m
Presto!
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I got
a = 0.30 m/s^2
Find Fax = 155.88 N and Fay = 72.69 N
Find Fn = 319.31 N
Find Ffr = 143.69
a = 0.305 m/s^2