Solve the following problem and then upload it. A horse pulls a cart up a hill at an angle of 53 degrees. The horse pulls with a force of 2200 N parallel to the ground. The cart has a mass of 1340 kg.

Question

Solve the following problem and then upload it. A horse pulls a cart up a hill at an angle of 53 degrees. The horse pulls with a force of 2200 N parallel to the ground. The cart has a mass of 1340 kg.

a. If the coefficient of friction between the cart and the ground is 0.43, what acceleration does the cart experience?

b. If it takes 50 s to pull the cart up the ramp starting from rest, how high is the hill? 

Answer 1

Let

α = 53º be the angle between the hill's ground and the horizontal,

F_h = 2200N be the force of the horse,

m = 1349 kg be the mass of the cart,

f = 0.43 be the coefficient of friction,

g = 9.8 m/s² be gravitational acceleration,

a be the cart's acceleration (to be computed)

QUESTION a.

First consider the force mg of gravity acting on the cart.

(At this point you need to draw a picture.)

Gravity, acting vertically down,

is the vector sum of a force F₁ parallel to the ground, acting downhill,

plus a force F₀ perpendicular to the ground.

They can be computed from one of the right angle triangles in your picture:

F₀ = m g cos(α)

F₁ = m g sig(α)

The force F₁ is the portion of gravity acting against the horse.

The force F₀ is the portion of gravity causing friction, acting against any move.

The force of friction

F_f = F₀ f

The acceleration a can be computed from Newton's second law

F = a m

where F is the net force acting on the cart.

To compute the net force F, we need to consider several situations.

While F₁ always acts downward against the force F_h of the horse,

F_f acts against any movement, up or down.

If F_h > F₁ then the horse is stronger than gravity F₁, the cart would move up,

and the force of friction F_f would act against the force of the horse:

F = F_h - F₁ - F_f

In contrast, if F_h < F₁ then gravity is winning, the cart would slide downhill, and

friction would be helping the horse against gravity:

F = F_h - F₁ + F_f

The third case is F_h = F_1.

That means that the horse and gravity are exactly balanced,

and there is no movement, with friction playing no role.

Each of the first two above cases divides into two more sub-cases.

Thus in total we have these cases:

Case 1: F_h > F₁

Then F = F_h - F₁ - F_f

Case 1.1: F > 0

That means that the horse is strong enough to overcome gravity as well as friction,

and the cart will move up.

Case 1.2: F <= 0

That mean that the horse is stronger than gravity but not strong enough to overcome friction,

and the cart will not move.

Case 2: F_h < F₁

Then F = F_h - F₁ + F_f

Case 2.1 F < 0

That means that gravity is strong enough to overcome the force of the horse as well as friction,

and the cart slides down.

Case 2.1 F >= 0

That mean that gravity is stronger than the horse, but not strong enough to overcome friction,

and the cart will not move.

Case 3: F_h = F₁

Then there is no movement.

Before proceeding let's figure out which is the case, given your numbers:

F₁ = m g sig(α)

= 1340 kg . 9.8 m/s² sig(53º) = 10487N

So we have the case 2 of F_h < F₁, where gravity is stronger than the horse. And hence

F = F_h - F₁ + F_f

F_f = m g cos(α) f

= 1340 kg . 9.8 m/s² cos(53º) 0.43 = 3398 N

F = F_h - F₁ + F_f

= 2200 - 10487 + 3398 = -4888 N < 0

Therefore we have case 2.1 -- the cart slides downhill under the net force F.

The acceleration

a = F/m

= -4888N/1340kg = -3.6 m/s²

The negative sign means that the acceleration is downhill.

QUESTION b.

From the answer to question a. we know that the cart will never reach

the top of the hill, contradicting the premise Question b.

That means that there is probably some mistake in the given numbers.

We could have seen it at the outset, just by considering the situation:

2200N is the weight of about 220 kg trying to pull up a counter-weight

six times larger on an extremely steep hill.

The poor horse would not even have a change on level ground against the force of friction.

Therefore I will answer this Question b. assuming some positive (uphill) acceleration a.

Let

t = 50s be the time when the cart reaches the top, starting at time 0

a be the cart's uphill acceleration (assumed already computed)

s be the length of the hill (unknown)

h be the height of the hill (to be computed).

By definition of acceleration

s = a t² / 2

From the right-angle triangle forming the hill:

h = s sin(α)

= a 50² sin(53º) / 2 = 998a