Technician A says inspecting the operation of the automatic emergency brake system helps determine whether the spring brakes wil
1 answer:
You might be interested in
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
Weight transfer is given as follows;
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
× 1.3 = 1.7 ×
+
= 18000
= 18000*17/30 = 10200 N
= 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
The least coefficient of friction, μ = 0.625.
Answer:
the magnitude of F_A is 752 N
the direction theta of F_A is 57.9°
Explanations:
Given that,
Resultant force = 1330 N in x direction
∑Fx = R
from the diagram of the question which i uploaded along with this answer
FB = 800 N
FAsin∅ + FBcos30 = 1330 N
FAsin∅ = 1330 - (800 × cos30)
FA = 637.18 / sin∅
Now ∑Fx = 0
FAcos∅ - FBsin30 = 0
we substitute for FA
(637.18 / sin∅)cos∅ = 800 × sin30
637.18 / 800 × sin30 = sin∅/cos∅
and we know that { sin∅/cos∅ = tan∅)
so tan∅ = 1.59295
∅ = 57.88° ≈ 57.9°
THEREFORE FROM THE EQUATION
FA = 637.18 / sin∅
we substitute ∅
so FA = 637.18 / sin57.88
FA = 752 N
