Newton's Laws of Motion

Question

A contestant in a winter games event pushes a 43.0 kg block of ice across a frozen lake as shown in the figure below.

A person is pushing a block of ice to the right on the surface of a frozen lake. The person is applying force on the top-left corner of the block. The arrow representing the force vector F is shown to act diagonally downward from the shoulder of the person to the top-left corner of the block. This arrow is at an angle of 25° below the horizontal from the person's shoulder level.

The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. (For each answer, enter a number.)

Calculate the minimum force F (in N) he must exert to get the block moving.

 __ N

What is its acceleration (in m/s²) once it starts to move, if that force is maintained?

 __m/s²

Answer 1

Let

m = 43 kg be the mass of the ice block,

α = 25° be the angle of the force F,

H = Fcos(α) be the horizontal component of the force F,

V = Fsin(α) be the vertical component of the force F,

s = 0.1 be the coefficient of static friction,

d = 0.03 be the coefficient of dynamic friction,

g = 9.81 m/s² be gravitational acceleration,

a (to be calculated) be the acceleration of the block once it starts moving.

Friction is due to the weight, mg, of the block plus the additional vertical force V.

Therefore the total force of static friction is

(mg + V)s

The minimal horizontal force, H, to overcome the force of friction must satisfy

H = (mg + V)s

Substituting H and V

Fcos(α) = (mg + Fsin(α))s

Solving for F

F = mgs/(cos(α) - sin(α)s) (1)

Substituting actual numbers

F = 43×9.81×0.1/(cos(25°) - sin(25°)×0.1) = 48.82 N

Now for the acceleration:

Like the force of static friction, the total force of dynamic friction is

(mg + V)d

So the net horizontal force acting on the block of ice is

H - (mg + V)d

By Newton's second law

H - (mg + V)d = ma

Solving for a

a = (H - (mg + V)d)/m = (H - Vd)/m - gd

Substituting H and V

a = (Fcos(α) - Fsin(α)d)/m - gd = (cos(α) - sin(α)d)F/m - gd (2)

At this point you could substitute the computed F = 48.82 N and get the acceleration a.

But I will substitute the symbolic solution (1) to illustrate the point that the

acceleration a does not depend on the mass m.

Substituting (1) into (2)

a = ((cos(α) - sin(α)d)mgs/(cos(α) - sin(α)s)) / m - gd

= gs(cos(α) - sin(α)d)/(cos(α) - sin(α)s) - gd

Notice that the minimal force F to overcome static friction does depend on the mass m,

but the acceleration a does not.

Substituting actual numbers

a = 9.81×0.1×(cos(25°) - sin(25°)×0.03)/(cos(25°) - sin(25°)×0.1) - 9.81×0.03 = 0.72 m/s²