Two forces, one of which double the other has resultant of 280N. if the direction of the large force is reversed and the remaini
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Answer:
v = 21.409 m/s
Explanation:
Given data:
Total weight of the platform and weight together, W = 900 N
Diameter of the water jet = 5 cm = 0.05 m
Now,
the force exerted by the water jet is balancing the platform
also,
Force exerted by the water jet = ρAv²
where,
ρ is the density of the water = 1000 kg/m³
A is the area of the outlet of the jet =
or
A = 0.00196 m²
v is the velocity of the jet
thus,
W = ρAv²
or
900 = 1000 × 0.00196 × v²
or
v = 21.409 m/s
Hence, the velocity of the jet is 21.409 m/s
Solution :
Finding the cohesion of the soil(c) using the relation:
Here, is the unconfined compression strength of the soil;
= 400 psf
∴ The cohesion value is greater than 0
So the use of the angle of internal friction is 0
Referring to the table relation between bearing capacity factors and angle of internal friction.
For the angle of inter friction
Therefore,
= 2386 psf
∴ Allowable bearing capacity
∴
∴ B = 0.2 ft
Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft