A vehicle with a mass of 500 kg rolls down a slanted road with an acceleration of 0.04 m / s2. The frictional force
1 answer:
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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