drowning_pocoyo
Member
- Joined
- Apr 15, 2009
- Messages
- 74
- Gender
- Male
- HSC
- 2010
Hey guys i need help with this motion question, i did it but im not entirely sure im correct:
A car of mass 2000kg negotiates a turn radius 130m, banked at an angle 9 degrees to the horizontal at a speed of 70km/h. Calculate the frictional force to the nearest newton. (Take acceleration to be 10 m/s)
Heres what i did:
Fnetx = F - mg sin 9, where F = frictional force
.: (mv^2)/r = F - mg sin 9 ---> (1)
Fnety = N - mg cos 9, where N = normal force
Since, the car is not rising or falling
.: 0 = N - mg cos 9
.: N = mg cos 9
From (1),
F = m[ (v^2)/r + g sin 9]
.: F = 2000 {[(70/3.6)^2]/130 + 10 sin 9}
.: F = 70959 N
thanks for the help in advance ;D
A car of mass 2000kg negotiates a turn radius 130m, banked at an angle 9 degrees to the horizontal at a speed of 70km/h. Calculate the frictional force to the nearest newton. (Take acceleration to be 10 m/s)
Heres what i did:
Fnetx = F - mg sin 9, where F = frictional force
.: (mv^2)/r = F - mg sin 9 ---> (1)
Fnety = N - mg cos 9, where N = normal force
Since, the car is not rising or falling
.: 0 = N - mg cos 9
.: N = mg cos 9
From (1),
F = m[ (v^2)/r + g sin 9]
.: F = 2000 {[(70/3.6)^2]/130 + 10 sin 9}
.: F = 70959 N
thanks for the help in advance ;D
Last edited:
