A vehicle 1200 kg of wheelbase 2.8 m is slowing down, a= -0.56 m/s2 to 54 km/h and dynamic load at front is 5790 N. Then, the vehicle accelerates to 95 km/h in 15 s. The height at centre of gravity of the vehicle is 0.5 m above the ground level. Take g=9.8 m/s2, find :
A. Load Transfer during deceleration
B. Length of x and y in meters
C. Dynamic load at rear axle during deceleration
D. Dynamic load at front and rear during acceleration
To find the answers to these questions, we can use some basic principles of physics such as Newton's laws of motion and the equation for load transfer.
A. Load Transfer during deceleration:
The load transfer during deceleration can be calculated using the equation:
Load Transfer = (mass of the vehicle) x (deceleration) x (height of the center of gravity) / (wheelbase)
Given:
Mass of the vehicle = 1200 kg
Deceleration = -0.56 m/s^2
Height of the center of gravity = 0.5 m
Wheelbase = 2.8 m
Load Transfer = (1200 kg) x (-0.56 m/s^2) x (0.5 m) / (2.8 m)
Load Transfer = -100 N
Therefore, the load transfer during deceleration is -100 N. The negative sign indicates that the load is transferred from the front axle to the rear axle.
B. Length of x and y in meters:
To find the length of x and y, we can use the equation for displacement:
Displacement = (initial velocity) x (time) + (1/2) x (acceleration) x (time)^2
Given:
Initial velocity = 54 km/h = 15 m/s (converted from km/h to m/s)
Time = 15 s (as mentioned in the question)
Acceleration = (change in velocity) / (time) = (95 km/h - 54 km/h) / 15 s = 41 km/h / 15 s = 11.39 m/s^2 (converted from km/h to m/s)
For x:
Displacement(x) = (15 m/s) x (15 s) + (1/2) x (11.39 m/s^2) x (15 s)^2
Displacement(x) = 225 m
For y:
Displacement(y) = (1/2) x (acceleration) x (time)^2
Displacement(y) = (1/2) x (11.39 m/s^2) x (15 s)^2
Displacement(y) = 1020.375 m
Therefore, the length of x is 225 meters and the length of y is 1020.375 meters.
C. Dynamic load at rear axle during deceleration:
The dynamic load at the rear axle during deceleration can be calculated using the equation:
Dynamic Load Rear = (mass of the vehicle) x (deceleration) - Load Transfer
Given:
Mass of the vehicle = 1200 kg
Deceleration = -0.56 m/s^2
Load Transfer = -100 N
Dynamic Load Rear = (1200 kg) x (-0.56 m/s^2) - (-100 N)
Dynamic Load Rear = 667.2 N - 100 N
Dynamic Load Rear = 567.2 N
Therefore, the dynamic load at the rear axle during deceleration is 567.2 N.
D. Dynamic load at front and rear during acceleration:
To find the dynamic load at the front and rear axles during acceleration, we need the total load on the axles.
Total Load = Load Transfer + static load
Given:
Static load at the front axle during deceleration = 5790 N
Static load at the front axle during acceleration can be calculated using the equation:
Static Load Front = (mass of the vehicle) x (acceleration) + Load Transfer + static load at the front axle during deceleration
Given:
Mass of the vehicle = 1200 kg
Acceleration = 11.39 m/s^2
Load Transfer = -100 N
Static Load Front = (1200 kg) x (11.39 m/s^2) + (-100 N) + (5790 N)
Static Load Front = 13668 N
Total Load = Static Load Front + Static Load Rear
Therefore, the dynamic load at the front axle during acceleration is 13668 N, and the dynamic load at the rear axle during acceleration can be calculated by subtracting the static load at the front axle during acceleration from the total load.
Dynamic Load Rear = Total Load - Static Load Front
Dynamic Load Rear = (13668 N + 567.2 N) - 13668 N
Dynamic Load Rear = 567.2 N
Therefore, the dynamic load at the rear axle during acceleration is 567.2 N, similar to the dynamic load during deceleration.