Answer :
Answer:
μ = 0.161
Explanation:
Since the tension on the rope (F) is being applied at an angle (θ), it should be decomposed into a vertical and a horizontal component.
The vertical component of the tension on the rope (F*sin(θ)) is pulling the sled upwards, decreasing the normal force.
Therefore, the normal force can be calculated as:
[tex]N = m*g - F*sin(\theta)\\N= 3*9.8 - 5*sin(30)\\N= 26.9 N[/tex]
The horizontal component of the tension on the rope (F*cos(θ)) is the force being applied to move the sled and that should equal the coefficient of friction multiplied by the normal force.
[tex]F*cos(\theta)=N*\mu\\5*cos(30)=26.9*\mu\\\mu=0.161[/tex]
the coefficient of friction between the sled and the ground is 0.161