A 125 kg crate rests on the flatbed of a truck that moves at a speed of 15.0 m/s around an unbanked curve whose radius is 66.0m. This crate does not slip relative to the truck. Obtain the magnitude of the static frictional force that the truck bed exerts on the crate.

Question

A 125 kg crate rests on the flatbed of a truck that moves at a speed of 15.0 m/s around an unbanked curve whose radius is 66.0m. This crate does not slip relative to the truck. Obtain the magnitude of the static frictional force that the truck bed exerts on the crate.

If you could explain/show how you got the answer to this, it would be great!
Thanks

Answer 1

The frictional force is compensating for the inward force to due to circular motion, for which you use the formula

F=(mv^2)/r
F=((125)(12^2))/66=273N
pretty sure that's correct

Answer 2

Thanks! That helped a lot

Answer 3

The pipe has mass of 500kg and rests on the flat bed of the truck. The static and dynamic coefficients of friction between the pipe and the truck bed are respectively μχ=0,04 andμκ=0,35 the truck starts from rest with a constant acceleration of

Answer 4

To find the magnitude of the static frictional force that the truck bed exerts on the crate, we can use the centripetal force equation. The centripetal force is the net force that acts towards the center of the circular path, keeping the crate in motion along the curved path.

Step 1: Identify the forces acting on the crate:
- The weight force acting downward: Fw = mg, where m is the mass of the crate (125 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The normal force acting upward, which is equal to the weight force: N = mg.

Step 2: Determine the centripetal force:
The centripetal force is given by the equation Fc = (mv^2) / r, where v is the velocity of the crate (15.0 m/s) and r is the radius of the curve (66.0 m).

Step 3: Equate the centripetal force to the sum of the horizontal forces:
The static frictional force, Fs, provides the necessary horizontal force to keep the crate moving in a circular path without slipping.

Since the crate does not slip relative to the truck, the static frictional force can be at its maximum value, given by Fs max = μs × N, where μs is the coefficient of static friction between the crate and the truck bed.

In this case, the static frictional force, Fs, is equal to the centripetal force, Fc.

So, we have Fs = Fc = (mv^2) / r.

Step 4: Solve the equation for the static frictional force:
Fs = (mv^2) / r = (125 kg)(15.0 m/s)^2 / 66.0 m

Calculating this expression, we get:
Fs ≈ 426.136 N

Therefore, the magnitude of the static frictional force that the truck bed exerts on the crate is approximately 426.136 N.