Matt Ashell, world famous downhill skier, waxed his skis with the wrong kind of wax and ended up with a relatively high coefficient of friction of 0.40. If Matt, who weights 667.5 N is on a downhill slope of 38 degrees, will he slide without a push?
mg = 667.5 N.
F1 = mg*sinA = 667.5*sin38 = 411 N = Force parallel to the hill.
F2 = mg*sinA= 667.5*cos38 = 526 N = Force perpendicular to the hill.
Fs = u*F2 = 0.4 * 526 = 210.4 N = Force of static friction.
Yes, because F1 is greater than the forceof s tatic friction, Fs.
Correction: F2 = mg*cosA.
To determine whether Matt will slide without a push, we need to compare the gravitational force pulling him downhill with the frictional force exerted by the skis.
First, let's calculate the gravitational force acting on Matt. We can use the formula:
Force (F) = mass (m) × acceleration due to gravity (g)
Here, the mass (m) is the weight (w) divided by the acceleration due to gravity (g ≈ 9.8 m/s²). Therefore:
m = w / g
Given that Matt's weight is 667.5 N, we have:
m = 667.5 N / 9.8 m/s²
Now we can calculate the mass:
m ≈ 68.1 kg
Next, we can calculate the component of the gravitational force acting parallel to the slope. This force is given by:
Force_parallel = m × g × sin(θ)
Where θ is the angle of the slope. Given that the slope angle is 38 degrees, we have:
Force_parallel = 68.1 kg × 9.8 m/s² × sin(38°)
Now we can calculate the frictional force using the coefficient of friction (μ) and the normal force (Force_perpendicular):
Force_friction = μ × Force_perpendicular
The normal force is the component of the gravitational force acting perpendicular to the slope, which is given by:
Force_perpendicular = m × g × cos(θ)
Putting it all together:
Force_friction = μ × m × g × cos(θ)
Given that the coefficient of friction (μ) is 0.40 and the slope angle (θ) is 38 degrees:
Force_friction = 0.40 × 68.1 kg × 9.8 m/s² × cos(38°)
Now let's calculate both the gravitational force parallel to the slope and the frictional force:
Force_parallel ≈ 421.4 N
Force_friction ≈ 269.8 N
Since the force of friction (269.8 N) is less than the parallel force component (421.4 N), Matt will slide without a push.