Force help please (friction too)?

Question

A truck with a mass of 4000kg is heading easy just before making a left turn on a flat, rough roadbed at a constant speed of 20 m/s. The radius of curvature of teh road is 100m

A- What is the direction and magnitude of the resultant force on the truck?

B- What is the monimum coefficient of static friction between the truck tires and the roadbed if the truck is not to slip?


HEEEEEELLLLLLLLLLLLLLLp!!!

Answer 1

> A- What is the direction and magnitude of the resultant force on the truck?

We know that F_net = ma (always!) And they give you "m", so then how much is "a"?

The answer is: EVERY time something is moving at a constant speed around a circle, its acceleration is:

a = v² / r

They give you "v" and "r". Do the math.

> B- What is the monimum coefficient of static friction between the truck tires and the roadbed if the truck is not to slip?

If the truck isn't slipping, it means that the force of friction must be great enough to keep the truck going in a circle. We already calculated how great that force needs to be (F_net, calculated above). So compare that with the max force of friction:

F_friction_max = (weight) × (μ)

Set F_net equal to F_friction_max, and solve for μ

Answer 2

A) I'm assuming the problem wants the force needed to accomplish the turn.

Centripetal force required to keep an object in a circle is mv^2/r. In this case that would be 4000*400/100= 16000 Newtons.

B) We need friction to supply 16000 newtons. Force of friction equals mu*mg = mu*4000*9.8. [mu = coefficient of friction.]

So, we need mu*4000*9.8 = 16000

so mu = .41