what is the force with which the curator pushes the sculpture?

Question

A museum curator pulls a 1470 N sculpture across the museum floor, with an acceleration of 0.5m/s2. If the coefficient of sliding friction between the floor and the sculpture is 0.57, what is the force with which the curator pushes the sculpture?

Answer 1

Define sculptor = person and sculpture = object.

Let's assume the person pushes the object to the right. This means the object experiences an applied force to the right and a frictional force to the left. Gravity is acting, so the object must experience a weight force pulling down. Since there is no vertical acceleration of the object, there must be a normal force from the floor pushing up on the object that is equal in magnitude to the gravitational force. Thus we can draw a free body diagram of the object with 4 forces acting on it.

Now we can write a Newton's 2nd Law equation for the horizontal direction.

F_A - F_F = ma

Solve for the applied force: F_A = ma + F_F

By definition of friction, F_F = µF_N

Plugging this frictional force into the equation for applied force gives:

F_A = ma + µF_N

We know that F_N = F_g because there is no vertical acceleration. Therefore

F_A = ma + µF_g

We are given a, µ, and F_g in the problem. So all we need now is the mass.

By definition, the weight of the object is F_g = mg. Therefore

m=F_g/g

Plug this into the equation:

F_A=(Fg/g)a + µF_g

Simplify:

F_A=F_g(a/g + µ)

We know that F_g = 1470 N, a=0.5 m/s², g=9.8 m/s², and µ=0.57.

Plug the numbers in and you will have your answer.