Workers have loaded a delivery truck in such a way that its center of gravity is only slightly forward of the rear axle, as shown in the drawing. The mass of the truck and its contents is 8340 kg. Find the magnitudes of the forces exerted by the ground on the following.
a) the front wheels (in N)
b)the rear wheels of the truck (in N)
To find the magnitudes of the forces exerted by the ground on the front and rear wheels, we need to consider the equilibrium of forces acting on the truck.
First, let's label the following variables:
- m = mass of the truck and its contents = 8340 kg
- g = acceleration due to gravity = 9.8 m/s^2
- d = distance between the center of gravity and the rear axle
- L = length of the truck from the rear axle to the front wheels
- F_front = force exerted by the ground on the front wheels
- F_rear = force exerted by the ground on the rear wheels
Since the center of gravity is only slightly forward of the rear axle, we can assume that the truck is at rest and in equilibrium. This means that the sum of the moments (torques) about the rear axle is zero.
The moment exerted by the weight of the truck and its contents about the rear axle is given by the formula:
Moment = (m * g * d) - (m * g * L/2)
Since the truck is in equilibrium, this moment must be equal to zero:
(m * g * d) - (m * g * L/2) = 0
We can rearrange this equation to solve for d:
(m * g * d) = (m * g * L/2)
d = L/2
Now that we know d = L/2, we can solve for the magnitudes of the forces:
a) the front wheels:
Since the truck is in equilibrium, the sum of the vertical forces must be zero.
F_front + F_rear = m * g
F_front = m * g - F_rear
b) the rear wheels of the truck:
Since the truck is in equilibrium, the sum of the moments about the rear axle must be zero.
(m * g * d) - (m * g * L/2) = 0
(m * g * L/2) = (m * g * d)
F_rear * L = m * g * (L/2)
F_rear = (m * g * (L/2)) / L
F_rear = m * g / 2
Now we can substitute the given values into the equations to find the magnitudes of the forces:
a) the front wheels:
F_front = (8340 kg * 9.8 m/s^2) - F_rear
b) the rear wheels of the truck:
F_rear = (8340 kg * 9.8 m/s^2) / 2
Compute the values to find the magnitudes of the forces exerted by the ground on the front and rear wheels.
To find the magnitudes of the forces exerted by the ground on the front and rear wheels of the truck, we need to analyze the forces acting on the truck.
We can start by considering the forces that affect the truck's vertical equilibrium. Since the truck is not accelerating vertically, the sum of the vertical forces must be zero.
Let's denote the mass of the truck and its contents as m, which is given as 8340 kg. Taking the acceleration due to gravity as 9.8 m/s^2, the weight of the truck is:
Weight = m * g = 8340 kg * 9.8 m/s^2 = 81732 N.
Now, let's consider the distribution of weight on the front and rear axles of the truck. Since the center of gravity is slightly forward of the rear axle, we can assume that a fraction of the truck's weight is acting on the rear wheels, and the remaining weight is acting on the front wheels.
Let's denote the fraction of weight acting on the front wheels as f. Therefore, the weight acting on the front wheels is f * Weight, and the weight acting on the rear wheels is (1 - f) * Weight.
To find the value of f, we need some more information or assumptions about the positioning of the center of gravity. Without this information, we won't be able to precisely determine the magnitudes of the forces.
Please provide any additional details about the positioning of the center of gravity or any assumptions we can make to proceed further.
Use the law of moments to answer this question.
The total of the rear wheel forces and the front wheel forces equals the weight, M g = 81,732 N.
Most of the weight will be on the rear wheels.
You will need to know the distance of the center of gravity from the rear wheels, and the distance between front and back wheels.
These numbers should be on the drawing that you have not provided.