This problem has three questions I believe:
>
How hard does the floor push on the crate?
<span>We have to find the net
vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N
The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
=
1721.63 N</span>
> Find the friction
force on the crate
<span>We
have to look for the net horizontal force Fh which results from Fp and Fg.
Since Fg is a normal force entirely, so we can say that the
horizontal component is zero:
Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
=
811.66 N</span>
> What is the minimum
coefficient of static friction needed to prevent the crate from slipping on the
floor?
We just need to compute the
ratio Fh to Fn to get the minimum μs.
μs = Fh / Fn
= 811.66 N / 1721.63 N
<span>=
0.47</span>
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then

From here, we clear the value of d.
We have then:

Rewriting:
Answer:
The magnitude of the car's displacement is:
d = 20 miles
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :



v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.