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A vehicle with a mass of 500 kg rolls down a slanted road with an acceleration of 0.04 m / s2. The frictional force
between the wheels of the vehicle and the road is 1,800 Newtons.
a. Sketch the situation.
b. What is the angle of elevation of the road?
c. The steepness of a road is frequently measured as grade, which expresses the slope of a hill as a ratio of the
change in height to the change in horizontal distance. What is the grade of the hill described in this problem?

Respuesta :

Answer:

a) Sketch  in annex

b)  angle is  22 °

c) grade of the hill is  tan 22°  =  0,40

Step-by-step explanation: See Annex for free body diagram

According to Newton´s second law

∑ Fy  =  0      ⇒   mg cos ∠CAB - Fn  = 0

Notice  ∠ CAB  =  ∠ POD  ( They have perpendicular sides )

mg  =  500 Kg * 9.8 m/sec²     ⇒   mg  =   4900 [N]

mg cos ∠CAB - Fn  = 0      ⇒  4900*cos ∠CAB   =  Fn      (1)

∑ Fx  =  ma

mg* sin  ∠CAB  - Fr  =  500* 0.04 m/sec²   =  20 [N]

4900 * sin  ∠CAB   =  1800  + 20  [N]

sin  ∠CAB  = 1820 / 4900   ⇒  sin  ∠CAB  = 0.3714

From tables we get

∠CAB  = 22°

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