Answer:
a) [tex]F = 13166.468\,N[/tex], b) [tex]t = 2\,s.[/tex]
Explanation:
a) The equations of equilibrium for the truck are: (x' is the axis parallel to ramp, y' is the axis perpendicular to ramp)
During braking:
[tex]\Sigma F_{x'} = - m\cdot g \cdot \sin \theta - F = m\cdot a[/tex]
[tex]\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0[/tex]
The deceleration experimented by the truck is:
[tex]a=\frac{(36\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )-(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )}{3\,s}[/tex]
[tex]a = -5\,\frac{m}{s^{2}}[/tex]
The braking force is:
[tex]F= - m\cdot (g\cdot \sin \theta+a)[/tex]
[tex]F = -(8000\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot \sin 20^{\textdegree}-5\,\frac{m}{s^{2}} ][/tex]
[tex]F = 13166.468\,N[/tex]
b) The additional time required for the truck to stop is:
[tex]t = \frac{0\,\frac{m}{s}-(36\,\frac{km}{h} )\cdot(\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} ) }{-5\,\frac{m}{s^{2}} }[/tex]
[tex]t = 2\,s.[/tex]
