An object with a mass of #6 kg# is on a plane with an incline of # - pi/6 #. If it takes #18 N# to start pushing the object down the plane and #1 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Question

An object with a mass of #6  kg# is on a plane with an incline of # - pi/6  #.  If it takes #18  N# to start pushing the object down the plane and #1  N# to keep pushing it, what are the coefficients of static and kinetic friction?

Nathan Ashcroft · Accepted Answer

#mu_s =( 3sqrt3)/10#
# mu_k = 1/(20sqrt6)# #f = mu N# #N = mgcos theta = mg sqrt3/2# Now that we have our prerequisites, let's start by thinking about the static case. #18 = mu mgsqrt3/2# #18 = mu 4 * 10 sqrt3/2# When we solve, we #mu_s =( 3sqrt3)/10# Examining the kinetic scenario first #1 = mu_k * 4*10*sqrt 3/2# When we solve, we obtain # mu_k = 1/(20sqrt6)#

Elena Albarran · Answer

undefined The coefficient of static friction is μs = 0.577 and the coefficient of kinetic friction is μk = 0.100.

Sebastian Acosta · Answer

undefined The force required to start pushing the object down the plane is the force of static friction, $ F_s $. The force required to keep pushing it once it's in motion is the force of kinetic friction, $ F_k $.

The force of static friction is given by:

$$ F_s = \mu_s \cdot N $$

where $ \mu_s $ is the coefficient of static friction and $ N $ is the normal force.

The normal force $ N $ is equal to the gravitational force acting perpendicular to the plane, which is $ mg \cos(\theta) $, where $ m $ is the mass of the object, $ g $ is the acceleration due to gravity, and $ \theta $ is the angle of the incline.

Given:
- $ m = 6 $ kg
- $ g \approx 9.8 $ m/s² (acceleration due to gravity)
- $ \theta = -\frac{\pi}{6} $

We have:
$$ N = mg \cos(\theta) = 6 \times 9.8 \times \cos\left(-\frac{\pi}{6}\right) $$

The force of kinetic friction $ F_k $ is given by:

$$ F_k = \mu_k \cdot N $$

where $ \mu_k $ is the coefficient of kinetic friction.

Given that it takes 18 N to start pushing the object down the plane and 1 N to keep pushing it, we have:

$$ F_s = 18 \text{ N} $$
$$ F_k = 1 \text{ N} $$

We can now solve for $ \mu_s $ and $ \mu_k $ using the formulas for static and kinetic friction.

$$ \mu_s = \frac{F_s}{N} $$
$$ \mu_k = \frac{F_k}{N} $$

After calculating $ N $, $ \mu_s $, and $ \mu_k $, we'll have the coefficients of static and kinetic friction.