Question

A box is sitting on a board. The coefficient of static friction between the box and the board is 0.830216. The coefficient of kinetic friction between the box and the board is 0.326245. One side of the board is raised until the box starts sliding. Give a variable legend for this problem.

a) What is the angle at which the box starts sliding? The model for this problem:
θ=__________________________________ Answer________________________________

b) What is the magnitude of its acceleration after it starts sliding? The model for this problem:
a=__________________________________ Answer________________________________

Answer 1

Step-by-step explanation:

Coefficient of static friction μs = .830216

Coefficient of kinetic friction μk = .326245

a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle

tanθ = μs

tanθ = .830216

θ = 39.7°

b )

When the box starts sliding , kinetic friction will be acting on it .

frictional force on the box = μk mg cos 39.7

net force on the box

= mg sin39.7 - μk mg cos 39.7

Applying Newton's law of motion

mg sin39.7 - μk mg cos 39.7 = m a

a = g sin39.7 - μk g cos 39.7

= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7

= 6.26 - 2.46

= 3.8 m /s² .