All the weight of the wooden board rests solely on the support situated at the center of the rod, while the other support positioned at one end experiences no reaction force, resulting in a 0 reaction force.
Thus, the reaction force at the central support corresponds to the weight of the board, whereas the end support has 0 reaction force.
Answer:
17.35 × 10^(-6) m
Explanation:
Mass; m = 50 kg
Weight; W = 554 N
From the formula:
W = mg
This simplifies to; 554 = 50g
g = 554/50
g = 11.08 m/s²
Also, using the formula;
mg = GMm/r²
hence; g = GM/r²
Rearranging gives;
r = √(GM/g)
With G as a known constant of 6.67 × 10^(-11) Nm²/kg²
r = √(6.67 × 10^(-11) × 50/11.08)
r = 17.35 × 10^(-6) m
Let's consider a few possibilities.
1. The lowest velocity of the paratrooper would be just before hitting the ground.
2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping.
Hence, we will convert 100 mi/h to ft/s:
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec.
Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to
8 s * 146.67 ft/s = 1173.36 ft.
This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible.
Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.
For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26
For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
