1. ## Circular Motion Problem

Ok,

A car of mass 1.5 tonnes pass over a bridge formed the arc of a circle of radius 20m.

a) Find the force exerted by the car on the road at the top of the bridge if the car is travelling at 10m/s.

b) What speed would cause the car to be on the point of lifting up off the bridge at its highest point

What I've done:

F = mv^2 / r
F = 1500 * 10^2 / 20
F = 7500 N
---- Don't know if this is anything (the answer in the book is 7200N)

tan(t) = v^2/rg
tan(t) = 10^2/20*9.8
tan(t) = 0.510
(t) = tan^-1(0.510)
(t) = 27.02 deg

Force of car on road = mg sin (t)
= 1500*9.8sin (27.02)
= 6678.23 N

However as I said previously the answer is 7200 N.

b) No idea really... not sure of the formula to use

The answer for b) is 14m/s.

2. Originally Posted by jeta
Ok,

A car of mass 1.5 tonnes pass over a bridge formed the arc of a circle of radius 20m.

a) Find the force exerted by the car on the road at the top of the bridge if the car is travelling at 10m/s.

b) What speed would cause the car to be on the point of lifting up off the bridge at its highest point

What I've done:

F = mv^2 / r
F = 1500 * 10^2 / 20
F = 7500 N
---- Don't know if this is anything (the answer in the book is 7200N)

tan(t) = v^2/rg
tan(t) = 10^2/20*9.8
tan(t) = 0.510
(t) = tan^-1(0.510)
(t) = 27.02 deg

Force of car on road = mg sin (t)
= 1500*9.8sin (27.02)
= 6678.23 N

However as I said previously the answer is 7200 N.

b) No idea really... not sure of the formula to use

The answer for b) is 14m/s.
a) Fnet = mg - N = 14700 - N

where N is the normal reaction force of the road on the car and the downwards direction is positive.

But Fnet = mv^2/r = 7500.

Therefore 7500 = 14700 - N => N = 7200 Newton.

From Newton III, size of force exerted by car on road = 7200 Newton.

b) When the car is on the point of "lifting up off the bridge" N = 0.

Therefore mv^2/r = mg

=> 1500 v^2/20 = 14700 => v = ....

3. Thanks so much... How do you just know everything about maths? You really do...

So yeah, thanks again.