An object with a mass of #6 "kg"# is on a ramp at an incline of #pi/8 #. If the object is being pushed up the ramp with a force of # 2 "N"#, what is the minimum coefficient of static friction needed for the object to remain stationary?

Question

An object with a mass of #6  "kg"# is on a ramp at an incline of #pi/8 #.  If the object is being pushed up the ramp with a force of # 2  "N"#, what is the minimum coefficient of static friction needed for the object to remain stationary?

William Audy · Accepted Answer

#mu_"s" >= 0.377#

First, sketch out a free body diagram.

Draw an object on a ramp inclined at an angle of #theta=pi/8#. Then draw the forces. There should be 4 forces. Namely, the weight #W = mg# (which points downwards), the normal #N# (which is normal to the ramp), the applied force, #F_"a"=2 "N"# (which points parallel to the slope upwards) and the static friction #F#.

The friction acts parallel to the slope but it is ok at this point to know whether it points upwards or downwards. Whether it points up the ramp or down the ramp depends on how #F_"a"# compares with #W#. If the #F_"a"# is too large, then #F# will point downwards, and vice versa. We take #F# to be positive if it points upwards and negative if it points downwards, but for know we don't know yet.

Apply the equilibrium condition in step two.

From Newton's Second Law (#F_"Net"=ma#), if an object is stationary, the net force acting on it must sum up to zero. This is a 2D problem so most likely a vector sum of the forces will result in two equations that you will need to solve.

Step 3: Selecting an intelligent coordinate scheme.

One way to perform vector addition is to resolve all the vectors into their individual components first, then add up the resolved vectors. Rather than using the convention that #x# is horizontal and #y# is vertical, we will use a coordinate system with #x# being parallel to the ramp, upwards being positive, and #y# being normal to the ramp, upwards being positive. This is because 3 out of 4 of the forces lies on either of these axis.

More specifically, it makes no difference whatsoever that we arbitrary designate the origin as the bottom of the ramp.

Step 4: The forces are added.

To achieve equilibrium, both the #x# and #y# component of the net force must be zero. For this part, you have to constantly refer the FBD that you have drawn previously. For the #x# component to be zero, we have

#F_"a" - mgsintheta + F = 0#.

Consequently,

#F = mgsintheta - F_"a"# #~~ 20.5 "N" > 0#.

The static friction required to hold the object points up the ramp. Now for the #y# component to be zero,

#N - mgcostheta = 0#.

Consequently,

#N = mg costheta#.

Step 5: Calculating the static friction minimum coefficient.

For the item to stay in place,

#F < mu_"s" N#,

where #mu_"s"# is the coefficient of static friction. Substituting the expression of the forces found in part 4,

#mgsintheta - F_"a" <= mu_"s" (mg costheta)#.

reorganizing yields

#mu_"s" >= frac{sintheta - frac{F_"a"}{mg}}{costheta} ~~ 0.377#.

Roughly 0.377 is the lowest coefficient of static friction.

Adrian Ayala · Answer

undefined The minimum coefficient of static friction needed for the object to remain stationary on the ramp is approximately 0.27.