# An object with a mass of #7 kg# is on a ramp at an incline of #pi/4 #. If the object is being pushed up the ramp with a force of # 8 N#, what is the minimum coefficient of static friction needed for the object to remain put?

##### 1 Answer

#### Answer:

#### Explanation:

As seen in the **free-body diagram**, there are four forces acting on the object.

- Gravitational Pull
#W = m * g = 68.67 color(white)(l) "N"# ; - Normal Force
#N# ; - Static Friction;
- The externally-applied force,
#color(purple)(F)# acting on the object upwards along the slope.

Consider forces acting on the object in two perpendicular directions: normal to the slope

Forces acting **normal** to the slope include

- the normal force,
#N_y = "N"# - a portion of the gravitation pull,
#W_y =W * cos(pi/4) = 48.56 color(white)(l) "N"#

The object experiences no acceleration on the direction normal to the slope as it stays in contact with the ramp. Thus the net force the object experiences on the direction normal to the slope shall equal zero.

#Sigma F_y = N_y -W_y =0#

#N_y = W_y = 48.56 color(white)(l) "N"#

Similarly, forces acting **along** the slope include

- the static friction,
#f_s# (#f# in the right-hand diagram) - a portion of the gravitational pull,
#W_x = W * sin(pi/4) = 48.56 color(white)(l) "N"# - the force seeking to push the object up along the ramp,
#color(purple)(F)= 8 color(white)(l) "N"#

Let the direction up along the slope be the positive direction of

The net force along the slope would also be zero since the box stays still and experiences no movement in that direction. Thus

#Sigma F = color(purple)(F_x) bb(color(navy)(+)) f_s - W_x = 0#

#8 color(white)(l) "N" + f_s = 48.56 color(white)(l) "N"#

#f_s = (48.56 -8) color(white)(l) "N"#

The static friction coefficient

#f_s le f_max = mu_s * N#

Rearranging gives

#mu_s = f_max / N ge f_s/N#

#mu_s ge f_s / N#

#color(white)(mu_s) = ((48.56 -8) color(white)(l) "N") / (48.56 color(white)(l) "N")#

#color(white)(mu_s) ~~ 0.835#

Therefore the minimum value of the static frictional coefficient would be

#mu_s ~~ 0.835#