I need help ASAP -Swimming at 2.50 m/s relative to still water, a swimmer heads

Question

I need help ASAP

-Swimming at 2.50 m/s relative to still water, a swimmer heads directly across a river of width 90 m. He arrives 25 m downstream from the point directly across the river from his starting point.

(a) What is the speed of the current in the river?

(b) At what angle relative to the line perpendicular to the shore should he head to arrive at a point directly opposite his starting point?

-A sled of mass m = 85.0 kg is being pulled horizontally on a frictionless floor by a force F = 200 N that makes an angle 45.0o above the horizontal. What is the magnitude of the normal force applied by the ground on the sled?

-In the figure we see two blocks connected by a string and tied to a wall, with ? = 34

Explanation / Answer

Let the width of the river be AB = 90 m
The point on the shore at which the swimmer arrives be C , such that BC = 25 m
AC = ? (90^2 + 25 ^2) = 93.41 m
Time taken by the swimmer to reach B in the absence of water current or in still water is
t = 90/ 2.5 = 36 s
In this time he is carried by the water current by a distance BC = 25 bm
Hence, the velocity of the water current = v = 30 / 36 = 0.833 m/s

a) Swimmer's speed w.r.t. the shore is the resultant of the velocity of the swimmer along AB and velocity of the water current along BC,
v' = ? (2.5^2 + 0.833^2 ) = 2.64 m/s
This also gives AC = v' x t = 95.04 m

b) The swimmer should head against the direction of the current making an angle ? with AB
sin ? = 25 / 95.04 = 0.507
? = 15.25 deg

Part B

If the box was sitting stationary (no force), the ground would exert a normal force of:
F = mg = 85*9.81 = 833.85 N

But the applied force partially lifts the sled. The sled is pulled upwards by
Fup = Fapp*sin(theta) = 200*sin(45)
Fup = 141.2N

Fn = 833.85 - 141.2 = 692.43N

Part C

The mass of the block A is not relevant to the question, that rope may as well be tied to the opposite wall.
So it's just a matter of the tension in the angled rope, when the mass is pulled sideways by the horizontal rope.
The angle is 32 degrees from horizontal. Mass B could be right at the join between the 2 ropes, that might making a FBD easier.
(1kg x g) = 9.8N. vertical force component (using 9.8 for g).
Tension in angled rope = (9.8/sin 32) = 18.49N.

Tension in horizontal rope = (9.8/tan 32) = 15.68N., incidentally.

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