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]
Answer:
Torque=13798.4 N.m
Explanation:
Given data
Mass of beam m₁=500 kg
Mass of the person m₂=70 kg
length of steel r₁=4.40m
center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m
To find
Torque
Solution
Torque due to beam own weight
Torque due to person
Now for total torque
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
<u>Where</u>:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
<u>Where:</u>
g: is gravity = 9.8 m/s²
Now, the horizontal velocity of the rock is:
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
So, the distance is:
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!