A jet transport has a weight of 1.87 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 12.0 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.
1 answer:
Answer:
a) R(fw) = 46.75*10⁶ (N)
b) R(rwi) = 70.125*10⁶ [N]
Explanation: See Attached file (the rectangle stands for a jet)
The diagram shows forces acting on the jet
Let R(fw) Reaction of front wheels and
R(rw) Reaction of rear wheels
Now we apply the Stevin relation, for R(fw) and a jet weight as follows
R(fw)/ 4 = 187*10⁶ / 16
Then :
R(fw) = ( 1/4) *187*10⁶ ⇒ R(fw) = 46.75*10⁶ (N)
And e do the same for the reaction on rear wheels
R(rw) / 12 = 187*10⁶ /16 ⇒ R(rw) =(3/4)*187*10⁶
R(rw) = 140,25*10⁶ [N]
The last expression is for the whole reaction, and must be devide by 2
because that force is exerted for two wheels, therefore on each of the two rear wheels the reaction will be:
R(rwi) = 70.125*10⁶ [N]
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Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Answer:
Farm = 98.1 [N]
Explanation:
To solve this problem we must draw the respective free body diagram, with the forces acting on the monkey. An analysis of the sums on the y-axis must be performed, in this axis the weight is acting down and the forces of both arms pulling up.
Weight is defined as the product of mass by gravitational acceleration.
W = m*g
where:
m = mass = 20 [kg]
g = gravity acceleration = 9.81 [m/s²]
W = 196.2 [N] (units of Newtons)
As this force points down, the force of both arms must go up, therefore each arm exerts a force of:
Farm = 196.2 / 2
Farm = 98.1 [N]
