An object with a mass of #14# # kg# is on a plane with an incline of # - pi/3 #. If it takes #12# # N# to start pushing the object down the plane and #7 # # N# to keep pushing it, what are the coefficients of static and kinetic friction?

Question

An object with a mass of #14# #  kg# is on a plane with an incline of # - pi/3  #.  If it takes #12# # N# to start pushing the object down the plane and #7 # #  N# to keep pushing it, what are the coefficients of static and kinetic friction?

Isabella Alvarez · Accepted Answer

#\mu_"static" = F_"frict"/(F_|_)=138.8/68.6=2.02#

#\mu_"kinetic" = F_"frict"/(F_|_)=125.8/68.6=1.83# Since the question doesn't specifically state it, we have to assume that once motion begins, it will always move at the same speed because that's the only explanation for the situation. There is a component of the weight force of the #14# #kg# mass that acts parallel to the inclined plane, which we can call #F_(|\|)#, and which tends to cause it to move down the slope. There is another component, #F_|_# that acts perpendicular to the slope, and dictates the magnitude of the frictional force. The frictional force itself acts up the slope and opposes movement down the slope. The sources of the forces will be: #F_(|\|) = mgsin heta=14xx9.8xxsin(-\pi/3)=-118.8# #N# #F_|_ = mgcos heta=14xx9.8xxcos(-\pi/3)=68.6# #N# For our purposes, we can disregard the negative sign in this instance because it simply indicates that the slope rises from right to left rather than from left to right. To calculate the coefficient of static friction, we know that there is a force of #118.8# #N# acting down the slope due to the weight of the object. It takes an additional #12# #N# to make it start to move, for a total of #138.8# #N#. This must just overcame - and therefore be virtually equal to - the force of static friction. The static friction expression is as follows: #F_"frict"=\mu_"stat"F_|_# Organizing: #\mu_"stat" = F_"frict"/(F_|_)=138.8/68.6=2.02# While a little unusual, a frictional coefficient of 2 is not unheard of. The argument for the coefficient of kinetic friction is the same, but the force that is required to maintain constant velocity versus the frictional force is #7# #N#. Combined with the component of the weight force acting in this direction, the total force is #125.8# #N#. #\mu_"kin" = F_"frict"/(F_|_)=125.8/68.6=1.83#

Axel Acevedo · Answer

undefined To find the coefficients of static and kinetic friction, we can use the equations:

1. $ F_{\text{friction, static}} = \mu_s \cdot N $
2. $ F_{\text{friction, kinetic}} = \mu_k \cdot N $

Where:
- $ F_{\text{friction, static}} $ is the static friction force,
- $ F_{\text{friction, kinetic}} $ is the kinetic friction force,
- $ \mu_s $ is the coefficient of static friction,
- $ \mu_k $ is the coefficient of kinetic friction, and
- $ N $ is the normal force acting on the object.

First, let's find the normal force:
$$ N = mg \cos(\theta) $$

Given:
- Mass, $ m = 14 \, \text{kg} $
- Incline angle, $ \theta = -\frac{\pi}{3} $
- Gravitational acceleration, $ g = 9.8 \, \text{m/s}^2 $

$$ N = (14 \, \text{kg}) \times (9.8 \, \text{m/s}^2) \times \cos\left(-\frac{\pi}{3}\right) $$

Now, we can use the given forces to find the coefficients of friction.

For static friction:
$$ F_{\text{friction, static}} = \mu_s \cdot N $$
$$ 12 \, \text{N} = \mu_s \cdot N $$

For kinetic friction:
$$ F_{\text{friction, kinetic}} = \mu_k \cdot N $$
$$ 7 \, \text{N} = \mu_k \cdot N $$

Dividing the equations:
$$ \frac{12 \, \text{N}}{7 \, \text{N}} = \frac{\mu_s \cdot N}{\mu_k \cdot N} $$
$$ \frac{12}{7} = \frac{\mu_s}{\mu_k} $$
$$ \mu_s = \frac{12}{7} \cdot \mu_k $$

Now, we need to find $ \mu_k $.
$$ \mu_k = \frac{7 \, \text{N}}{N} $$
$$ \mu_k = \frac{7 \, \text{N}}{14 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos\left(-\frac{\pi}{3}\right)} $$

Finally, we can substitute the value of $ \mu_k $ back into the equation for $ \mu_s $.

$$ \mu_s = \frac{12}{7} \cdot \frac{7 \, \text{N}}{14 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos\left(-\frac{\pi}{3}\right)} $$

Calculate the numerical values to find $ \mu_s $ and $ \mu_k $.