A jet transport has a weight of 2.03 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.1 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.
To determine the normal force exerted by the ground on the front wheel and each of the two rear wheels, we can make use of the principle of equilibrium. According to this principle, in order for an object to be in equilibrium, the sum of all the forces acting on it must be zero.
Let's start by determining the forces acting on the jet transport:
1. Weight (W) = 2.03 x 10^6 N: This is the force due to gravity acting vertically downward. It acts at the center of gravity of the plane.
Next, we need to determine the distances involved:
1. Distance between the front wheel and the rear wheels (L) = 16.0 m.
2. Distance between the front wheel and the center of gravity (d) = 12.1 m.
Now, let's solve the problem step by step:
(a) Normal force on the front wheel:
Since the plane is at rest on the runway, the vertical forces acting on it must balance out. The normal force exerted by the ground on the front wheel (Nf) must equal the weight of the plane (W).
Therefore, Nf = W = 2.03 x 10^6 N.
(b) Normal force on each of the two rear wheels:
Since the plane is at rest and in equilibrium, the sum of the moments about any point must be zero. We can choose the front wheel as the reference point.
The clockwise moment (or torque) due to the weight of the plane (W) acting at the center of gravity (12.1 m from the front wheel) must be balanced by the anticlockwise moment due to the normal force acting on the rear wheels.
The moments can be calculated using the formula: Moment = Force x Distance.
The moment due to the weight of the plane is given by: Moment(W) = W x d.
The moment due to the normal force on each rear wheel is given by: Moment(Nr) = 2 x Nr x L, as there are two rear wheels.
Since the plane is at rest, the sum of the moments must be zero.
Therefore, Moment(W) = Moment(Nr1) + Moment(Nr2), where Nr1 and Nr2 are the normal forces on each of the two rear wheels.
W x d = 2 x Nr x L.
We can rearrange this equation to solve for Nr:
Nr = (W x d) / (2 x L).
Substituting the given values, we have:
Nr = (2.03 x 10^6 N x 12.1 m) / (2 x 16.0 m).
Calculating this equation gives us:
Nr = 1.224 x 10^6 N.
Hence, the normal force exerted by the ground on each of the two rear wheels is 1.224 x 10^6 N.
To determine the normal force exerted by the ground on the front wheel and each of the two rear wheels of the jet transport, we can use the principle of moments.
(a) Normal force on the front wheel:
Step 1: Calculate the weight distribution of the jet transport.
The weight of the jet transport is given as 2.03 x 10^6 N. Since the plane is at rest, the weight is balanced by the normal forces on the front and rear wheels. Therefore, the weight is distributed between these two wheels.
Let Wf be the normal force on the front wheel, and Wr be the normal force on each rear wheel.
According to the principle of moments, the net moment about the rear wheels must be zero for the system to be in rotational equilibrium. Therefore, we can use the equation:
(Wf × 16.0 m) - (Wr × 12.1 m) = 0
Step 2: Calculate the weight distribution.
The weight distribution is given by the equation:
Wf + Wr + Wr = 2.03 x 10^6 N
Rearranging the equation, we get:
Wf + 2Wr = 2.03 x 10^6 N
Step 3: Substitute the value of Wr from Step 1 into the weight distribution equation.
Substituting (Wr × 12.1 m)/16.0 m for Wf in the weight distribution equation, we get:
(Wr × 12.1 m)/16.0 m + 2Wr = 2.03 x 10^6 N
Step 4: Solve the equation to find Wr.
Multiply through by 16.0 m to eliminate the denominator:
(Wr × 12.1 m) + 2Wr × 16.0 m = (2.03 x 10^6 N) × 16.0 m
Simplifying:
12.1Wr + 32.0Wr = 32.48 x 10^6 N
44.1Wr = 32.48 x 10^6 N
Wr = (32.48 x 10^6 N) / 44.1
Wr ≈ 737,810.20 N
Step 5: Calculate the normal force on the front wheel.
Using the equation from Step 3:
Wf ≈ (737,810.20 N × 12.1 m) / 16.0 m
Wf ≈ 556,381.36 N
Therefore, the normal force exerted by the ground on the front wheel is approximately 556,381.36 N.
(b) Normal force on each rear wheel:
The normal force on each rear wheel is equal to Wr, which we calculated in Step 4.
Therefore, the normal force exerted by the ground on each of the two rear wheels is approximately 737,810.20 N.