The weight of the van is 1750 Kgs and CG height of the van above the ground is 7
ID: 1860161 • Letter: T
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The weight of the van is 1750 Kgs and CG height of the van above the ground is 750 mm.style="font-size: 12.0pt; font-family: 'arial','sans-serif'; mso-fareast-font-family: 'times new roman'; mso-ansi-language: en-us; mso-fareast-language: en-us; mso-bidi-language: ar-sa;" data-mce-style="font-size: 12.0pt; font-family: 'arial','sans-serif'; mso-fareast-font-family: 'times new roman'; mso-ansi-language: en-us; mso-fareast-language: en-us; mso-bidi-language: ar-sa;">
Determine the deceleration rate during braking and braking efficiencies for this conditionstyle="font-size: 12.0pt; font-family: 'arial','sans-serif'; mso-fareast-font-family: 'times new roman'; mso-ansi-language: en-us; mso-fareast-language: en-us; mso-bidi-language: ar-sa;" data-mce-style="font-size: 12.0pt; font-family: 'arial','sans-serif'; mso-fareast-font-family: 'times new roman'; mso-ansi-language: en-us; mso-fareast-language: en-us; mso-bidi-language: ar-sa;">
Figure 2
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Determine the deceleration rate during braking and braking efficiencies for this condition The Van (See Figure 2) is braking with the forces shown developed at the front and rear axles. The weight of the van is 1750 Kgs and CG height of the van above the ground is 750 mm. Determine the deceleration rate during braking and braking efficiencies for this conditionExplanation / Answer
Nf and Nr be the normal reaction at the front and rare wheels. Both acting upwards
Ff and Fr be the friction forces t the front and rare wheels. Both acting towards right
let u be the friction coefficient
Ff = u*Nf-----------1
Fr = u*Nr-----------2
Force balance in x and y direction
Ff+ Fr = 4250+2500
Ff+ Fr = 6720 N-------3
Nf + Nr = 17500 N-------4
Moment balance about CG
Nr*0.75 -Nf*0.75+2500*1.8 -Fr*1.8 +4250*1.2 -Ff*1.2 = 0
Nr*0.75 -Nf*0.75+2500*1.8 -u*Nr*1.8 +4250*1.2 -u*Nf*1.2 = 0-----------------5
Substituting 1 and 2 in 3
u*(Nf+Nr) = 6750
from 4
u*17500 = 6750
u = 0.39
From 5
Nr*0.048 -Nf*1.218 +9600 = 0
Nf*1.218 -Nr*0.048 = 9600-----6
solving 4 and 6
Nf = 663.3 ad Nr = 16836.7
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