Answer:
[tex]F=39,68N[/tex]
Explanation:
Data:
Mass [tex]m=25 Kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.12[/tex]
Angle = [tex]29^{0}[/tex]
Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]
Solution:
By Newton's first law we know that for the x-axis:
[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.
And for the y-axis:
[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.
We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:
[tex]F=m.a[/tex]
[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .
Now, the horizontal component of the force in the rope will be given by
[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.
To obtain the friction force, we must know the normal force:
[tex]F_f=\mu. N[/tex]
Clearing N in the y-axis equation:
[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]
So we can express the x-axis equation as follows:
[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]
Finally, solving for F we get
[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]
[tex]F=39,68N[/tex]
