Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.
D, I hope this is correct!
Answer:
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Explanation:
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