Answer:
the tension in the cable= 929.33N
Explanation:
the tension in the cable is equivalent to the frictional force experienced between the road and the stunt man
coefficient of friction= μ=[tex]\frac{F_{r} }{N}[/tex]
μ=0.870
N= Normal force =m × g
m= 109kg
g= acceleration due to gravity =9.8m/s²
[tex]F_{r}[/tex]= μ × N
[tex]F_{r}[/tex]= μ × m × g
[tex]F_{r}[/tex]= 0.870 × 109 × 9.8
[tex]F_{r}[/tex]=929.33N
The tension in the cable is 929.33N
