A truck has wheels of diameter 0.9m, a weight of 8000kg, a wheel-base of 4m, and a tray height
of approximately 1.2m. The truck’s weight distribution is 50% on the rear wheels and 50% on the front wheels. The truck’s front springs have a combined stiffness of 550,000 N/m, and the rear
springs have a combined stiffness of 600,000 N/m. The front and rear spring lengths are both 0.75m
(uncompressed). The front and rear shock absorbers (dampers) both have damping values of 30,000 Ns/m. The elastic coefficient of restitution for contact of the wheels on the road is e = 0.5.
A 3000kg package (1m by 1m) is placed on the tray of the truck with its centre of gravity 1m in
front of the rear axle. The package is not restrained and is free to slide on the tray of the truck with
friction coefficient µs=µk=0.5. The truck then travels over a triangular speed bump 100mm high and 300mm wide.
(a) Determine the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray.
(b) Determine the maximum acceleration which the package experiences when crossing the
bump at this speed.
To solve this problem, we can break it down into several steps:
Step 1: Calculate the force on the package during the bump
Step 2: Calculate the maximum static friction force to prevent the package from sliding
Step 3: Calculate the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray
Step 4: Calculate the maximum acceleration the package will experience when crossing the bump at this maximum velocity
Let's go through each step in detail:
Step 1: Calculate the force on the package during the bump
The force on the package during the bump can be calculated using the concept of work-energy principle.
The work done on the package by the bump is equal to the change in its kinetic energy.
Work done = Change in kinetic energy
The work done by the bump is equal to the force applied by the bump multiplied by the distance over which the force is applied.
Work done = Force on the package * Distance over which the force is applied
Since the bump is triangular, the average force over the entire bump can be calculated as the maximum force applied by the bump divided by 2.
Average force = Maximum force / 2
Now, let's substitute the above equations:
Force on the package * Distance over which the force is applied = (Maximum force / 2) * Distance over which the force is applied
Force on the package = Maximum force / 2
The maximum force applied by the bump can be calculated using the weight of the truck and the package.
Maximum force = Weight of the truck + Weight of the package
Weight of the truck = 8000 kg * 9.8 m/s^2
Weight of the package = 3000 kg * 9.8 m/s^2
Step 2: Calculate the maximum static friction force
The maximum static friction force can be calculated using the equation:
Maximum static friction force = Coefficient of static friction * Normal force
The coefficient of static friction is given as 0.5.
The normal force can be calculated as the weight of the package.
Step 3: Calculate the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray
To calculate the maximum velocity, we equate the force on the package (Step 1) with the maximum static friction force (Step 2).
Force on the package = Maximum static friction force
Maximum force / 2 = Coefficient of static friction * Normal force
Rearranging the equation:
Maximum force = 2 * Coefficient of static friction * Normal force
Now we can substitute the values and solve for the maximum force.
Step 4: Calculate the maximum acceleration the package will experience when crossing the bump at this maximum velocity
The maximum acceleration can be calculated using the equation for acceleration due to a force:
Acceleration = Force on the package / (Weight of the package + Weight of the truck + Force on the package)
Now we can substitute the values and solve for the maximum acceleration.
Let's plug in the values and calculate the answers.
To determine the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray, we need to consider the forces and moments acting on the package.
(a) First, let's calculate the static friction force between the package and the tray. The weight of the package is 3000kg * 9.8m/s^2 = 29,400N. The weight distribution is 50% on the rear wheels, so the force on the rear wheels is 0.5 * 29,400N = 14,700N.
The maximum static friction force can be calculated using the formula: F_static_max = µ_s * N, where µ_s is the static friction coefficient and N is the normal force.
F_static_max = 0.5 * 14,700N = 7,350N
Since the package is 1m in front of the rear axle, there is a moment acting on the package due to the weight. The moment can be calculated as M = F_static_max * d, where d is the distance between the rear axle and the center of gravity of the package.
M = 7,350N * 1m = 7,350Nm
Now let's consider the forces and moments acting on the package when crossing the speed bump. The maximum force acting on the package is the weight of the package plus the additional force due to the acceleration caused by the upward slope of the speed bump.
The additional force due to the acceleration caused by the upward slope of the speed bump can be calculated using Newton's second law: F = m * a, where m is the mass of the package and a is the acceleration.
The mass of the package is 3000kg. The acceleration can be calculated using the distance traveled while crossing the speed bump and the time it takes to cross it. However, since we are interested in the maximum velocity at which the package does not move relative to the tray, we can assume that the acceleration is zero. Thus, the additional force is zero.
Therefore, the maximum force acting on the package is the weight of the package, which is 29,400N.
To determine the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray, we can apply the principle of conservation of energy. The energy gained by the truck and the package due to their vertical motion as they cross the speed bump must be equal to the energy lost due to the compression of the front and rear springs.
The energy gained by the truck and the package can be calculated as the work done against gravity while going up the speed bump:
Energy_gained = weight * height
Energy_gained = 29,400N * 0.1m = 2940J
The energy lost due to the compression of the front and rear springs can be calculated as the work done to compress the front and rear springs:
Energy_lost = (spring_constant * spring_compression^2) / 2
The spring constant for the front springs is 550,000 N/m, and the compression of the front springs can be calculated as the distance traveled by the truck while the front wheels are on the speed bump. Assuming a triangular speed bump, the distance traveled can be calculated as the hypotenuse of a right triangle with sides of 100mm and 300mm:
Distance_traveled = sqrt((0.1m)^2 + (0.3m)^2) = 0.316m
The spring compression for the front springs is the change in length, which is the uncompressed length of 0.75m minus the distance traveled:
Spring_compression_front = 0.75m - 0.316m = 0.434m
Now we can calculate the energy lost:
Energy_lost_front = (550,000 N/m * (0.434m)^2) / 2 = 67,780J
The energy lost for the rear springs can be calculated in the same way. The spring compression for the rear springs is also 0.434m. The spring constant for the rear springs is 600,000 N/m. Therefore:
Energy_lost_rear = (600,000 N/m * (0.434m)^2) / 2 = 76,170J
So the total energy lost due to the compression of the springs is:
Energy_lost = Energy_lost_front + Energy_lost_rear = 144,950J
Since the energy gained must be equal to the energy lost, we have:
Energy_gained = Energy_lost
2940J = 144,950J
To solve for the maximum velocity, we can use the formula for kinetic energy:
Kinetic_energy = (1/2) * mass * velocity^2
Using the mass of the truck and the package, which is 11,000kg (8,000kg for the truck + 3,000kg for the package), and rearranging the equation for velocity, we get:
velocity^2 = (2 * 2940J) / 11,000kg
velocity^2 = 0.533
velocity = sqrt(0.533) = 0.73 m/s (approximately)
Therefore, the maximum velocity at which the truck can cross the speed bump without the package moving relative to the tray is approximately 0.73 m/s.
(b) To determine the maximum acceleration experienced by the package when crossing the bump at this speed, we can use Newton's second law: F = m * a, where F is the maximum force acting on the package, m is the mass of the package, and a is the maximum acceleration.
The maximum force acting on the package is the weight of the package, which is 29,400N.
Therefore, the maximum acceleration experienced by the package is:
a = F / m = 29,400N / 3000kg = 9.8 m/s^2
So, the maximum acceleration experienced by the package when crossing the bump at the maximum velocity is 9.8 m/s^2.