Answer:
The coefficient of friction is 0.38.
Explanation:
The free body diagram is drawn below.
Let [tex]f[/tex] be frictional force acting in the backward direction as shown. Let the coefficient of friction be [tex]\mu[/tex]. Let [tex]N[/tex] be the normal reaction force acting on the bag.
Given:
Mass of the bag is, [tex]m=8.10\textrm{ kg}[/tex]
Force acting at [tex]\theta = 38[/tex]° is [tex]F= 29.5\textrm{ N}[/tex]
Acceleration due to gravity is, [tex]g=9.8\textrm{ }m/s^{2}[/tex]
The force F can be resolved into its components as [tex]F_{x}=F \cos \theta[/tex] and [tex]F_{y}=F \sin \theta[/tex]
Therefore,
[tex]F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}[/tex]
Now, as there is no acceleration in vertical direction, therefore,
Sum of upward forces = Sum of downward forces
[tex]N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}[/tex]
Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.
Therefore, backward force = forward force.
[tex]f=F_{x}\\f=23.25\textrm{ N}[/tex]
Now, frictional force is given as:
[tex]f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38[/tex]
Therefore, the coefficient of friction is 0.38.
