A dray horse is being pulled by a rope across a level plow field by a force of 750.0N exerted at an angle of 47° above the horizontal. If the horse’s velocity is constant and the coefficient of friction is ì=.21, determine the mass of the horse.
To determine the mass of the horse, we need to first analyze the forces acting on it.
Let's break down the given information:
Force exerted on the horse = 750.0 N
Angle above the horizontal = 47°
Coefficient of friction = μ = 0.21
The force exerted on the horse can be divided into two components:
1. The horizontal component (Fh) that opposes the force of friction.
2. The vertical component (Fv) that counters the weight of the horse.
We can find the horizontal component (Fh) using trigonometry:
Fh = Force × cos(Angle)
Fh = 750.0 N × cos(47°)
Next, let's determine the frictional force (Ffr):
Ffr = coefficient of friction × normal force
Since the horse is on a level plow field, the normal force (Fn) is equal to the weight (mg), where g is the acceleration due to gravity (9.8 m/s^2). Therefore:
Ffr = μ × Fn
Ffr = μ × (mg)
We can express the weight of the horse (mg) as the vertical component of the force exerted on the horse (Fv) by considering the angle:
Fv = Force × sin(Angle)
Fv = 750.0 N × sin(47°)
Since the force of friction (Ffr) opposes the horizontal component (Fh), we can equate them:
Fh = Ffr
750.0 N × cos(47°) = μ × (mg)
Furthermore, the vertical component (Fv) must be equal to the weight (mg):
Fv = mg
750.0 N × sin(47°) = mg
Now, we have a system of equations:
750.0 N × cos(47°) = μ × (mg)
750.0 N × sin(47°) = mg
To solve for the mass (m), we can divide both equations:
[tan(47°)] * sin(47°) = [cos(47°)] * [μ * (mg)] / mg
tan(47°) = μ
Plugging in the given value of the coefficient of friction (μ = 0.21), we get:
tan(47°) = 0.21
Using a scientific calculator to find the inverse tangent (arctan) of 0.21, we determine:
47° = arctan(0.21)
Therefore, the mass of the horse (m) can be calculated as:
m = 750.0 N × sin(47°) / (9.8 m/s^2)