Question

A 1.40 kg box rests on a plank that is inclined at an angle of 29° above the horizontal. The upper end of the box is attached to a spring with a force constant of 21 N/m, as shown in the figure. If the coefficient of static friction between the box and the plank is 0.80, what is the maximum amount the spring can be compressed and the box remain at rest?

Answer 1

By, using the equation F=kx we determine the force on the spring to be Fspring=21x since we are trying to find the displacement

the forces acting on the box must be broken down into their respective x and y component

since there is no movement on in the y direction

y: Fn-mgcostheta = 0

Fn = mg * costheta

Fn = (1.4)(9.8) * cos29º

Fn = 16.19

in the x direction we have

µsFn - mg sintheta = F

and since the system is in equilibrium we set that equal to the force of the spring and thus:

µsFn - mgsin theta = 23x

(.80*16.19) - (1.4)(9.8)(sin 29º) = 23x

12.9-6.5=23x

x= .28m