1. At the start of a race, the rear drive wheels B of the 1550 − 𝑙𝑏 car slip on the track.
Determine the car’s acceleration and the normal reaction the track exerts on the
front pair of wheels A and rear pair of wheels B. Also, determine the maximum
acceleration that can be achieved by the car without having the front wheels A leave
the track or the rear drive wheels B slip on the track. The coefficient of kinetic
friction is µ𝑘 = 0.7, and the mass center of the car is at G. The front wheels are free
to roll. Neglect the mass of all the wheels.
Given:
𝑊𝑐𝑎𝑟 = 1550 𝑙𝑏
µ𝑘 = 0.7
Required:
𝑁𝐴 , 𝑁𝐵 , 𝑎
𝑎𝑚𝑎𝑥 without having the front wheels A leave the track or the rear drive wheels B
slip on the track
Solution:
Equations of Motion: Since the rear wheels B are required to slip, the frictional force
developed is
𝐹𝐵 = µ𝑠 𝑁𝐵 = 0.7𝑁𝐵
1550
+← ∑𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥 ; 0.7𝑁𝐵 = 32.2 𝑎
+↑ ∑𝐹𝑥 = 𝑚(𝑎𝐺 )𝑦 ; 𝑁𝐴 + 𝑁𝐵 − 1550 = 0
↶ +∑𝑀𝐺 = 0;
𝑁𝐵 (4.75) − 0.7𝑁𝐵 (0.75) − 𝑁𝐴 (6) = 0
Solving the equations simultaneously,
𝑵𝑨 = 𝟔𝟒𝟎. 𝟒𝟔 𝒍𝒃 = 𝟔𝟒𝟎 𝒍𝒃
𝑵𝑩 = 𝟗𝟎𝟗. 𝟓𝟒 𝒍𝒃 = 𝟗𝟏𝟎 𝒍𝒃
𝒂 = 𝟏𝟑. 𝟐 𝒇𝒕/𝒔𝟐
Consider the FBD and the KD when the front wheels are about to leave the track,
𝑁𝐴 = 0, and the rear wheels B are not slipping,
Equations of Motion:
1550
+← ∑𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥 ; 𝐹𝐵 = 32.2 𝑎
+↑ ∑𝐹𝑥 = 𝑚(𝑎𝐺 )𝑦 ; 𝑁𝐴 + 𝑁𝐵 − 1550 = 0
↶ +∑𝑀𝐺 = 0;
𝑁𝐵 (4.75) − 𝐹𝐵 (0.75) − 𝑁𝐴 (6) = 0
If we assume that the front wheels are about to leave the track, 𝑁𝐴 = 0. Substituting
this value into our equations of motion,
𝑁𝐴 = 1550 𝑙𝑏
𝑁𝐵 = 9816.67 𝑙𝑏
𝑎 = 203.93 𝑓𝑡/𝑠 2
Since 𝐹𝐵 > (𝐹𝐵 )𝑚𝑎𝑥 = µ𝑠 𝑁𝐵 = 0.9(1550)𝑙𝑏 = 1395 𝑙𝑏, the rear wheels will slip.
Thus, the solution must be reworked so that the rear wheels are about to slip.
𝐹𝐵 = µ𝑠 𝑁𝐵 = 0.9𝑁𝐵
Solving our equations of motion together with our new equation for 𝐹𝐵
simultaneously,
𝑁𝐴 = 626.92 𝑙𝑏
𝑁𝐵 = 923.08 𝑙𝑏
𝒂𝒎𝒂𝒙 = 𝟏𝟕. 𝟐𝟔 𝒇𝒕/𝒔𝟐 = 𝟏𝟕. 𝟑 𝒇𝒕/𝒔𝟐