(a) 1800 N
The equation of the forces along the vertical direction is:
[tex]F sin \theta + N - mg = 0[/tex]
where
[tex]F sin \theta[/tex] is the component of the applied force along the vertical direction
N is the normal force on the sled
mg is the weight of the sled
Substituting:
F = 1210 N
m = 246 kg
[tex]\theta = 30.3^{\circ}[/tex]
We find N:
[tex]N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N[/tex]
(b) 0.580
The equation of the forces along the horizontal direction is:
[tex]F cos \theta - \mu_s N = 0[/tex]
where
[tex]F cos \theta[/tex] is the horizontal component of the push applied by the mule
[tex]\mu_s N[/tex] is the static frictional force
Substituting:
F = 1210 N
N = 1800 N
[tex]\theta = 30.3^{\circ}[/tex]
We find [tex]\mu_s[/tex], the coefficient of static friction:
[tex]\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580[/tex]
(c) 522 N
In this case, the force exerted by the mule is
[tex]F= 6.05 \cdot 10^2 N = 605 N[/tex]
So now the equation of the forces along the horizontal direction can be written as
[tex]F cos \theta - F_f = 0[/tex]
where
[tex]\theta=30.3^{\circ}[/tex]
and [tex]F_f[/tex] is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value [tex]\mu N[/tex], depending on how much force is applied in the opposite direction)
Solving the equation,
[tex]F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N[/tex]
