Question

03.2 A boat must cross a 190-m-wide river and arrive at a point 30 m upstream from where it starts (see figure). To do so, the pilot must head the boat at an angle ?         = 40

Answer 1

Hello,

3.2 The first step is to dram a picture. The boat needs to cross the 190 m wide river and go 30 m upstream.The y-direction is 190 and x-direction is 30m. Connect the starting point and the ending point with a line. Now you should have a right triangle drawn. We also know that the boat drove at a 40 degree angle upstream so add that info into the triangle.

The boat is going 4.2m/s from the starting point to the finishing point. Determine the x and y components of velocity by using the angle given. Vy = (cos40)4.2 = 3.2m/s Vx = (sin40)4.2 = 2.7m/s Now we know a current is pushing in the negative x direction against the boat, but it does not affect the y direction. Therefore, we can determine how long it will take the boat to get the 190m across the river going 3.2m/s. 190m/3.2m/s = 59 seconds  

Now since we know it took 59 seconds to get across the river we can determine the velocity in the x direction. 30 m/59 s = .508 m/s. And we know that the boat can go 2.7m/s, so 2.7 - .508 gives the velocity of the current which is 2.2 m/s

3.3 The boat can travel 8 km/hr, but it traveled 3km in 1 hr or 3km/hr, therefore 8km/hr - 3km/hr = 5km/hr which is the speed of the current.

The float was dropped and went downstream for 1 hr at 5 km/hr until the boat turned around, so the float was 5km from the dropped point. and the boat was 3 km in the opposite direction from the drop point. 5+3 = 8km. The boat is 8 km from the float when it turns around. Once the boat turns around the float is no longer gaing distance away from the boat and the boat can go 8km/hr faster than the float, therefore the boat will take 1 hr to reach the float once it turns around. And the boat will catch the float 10 km away from the drop point. The float is 5km away from the drop point when the boat turns around and then the float moves another 5km during the hour that the boat is catching up.

3.4 Draw the coordinate system with all the force vectors.

split the forces into x and y components and dont forget to use a negative sign where needed.

F1x = cos(25)160

F1y = sin(25)160

F2x = -cos(15)F2

F2y = -sin(15)F2

F3x =0

F3y = -F3

If stationary the sum of the forces are equal to 0 and the the object is in static equillibrium.

F1x -F2x = 0 F1x = F2x cos(25)160 = cos(15)F2 solve for F2 = 150

F1y -F2y - F3 = 0 sin(25)160 - sin(15)F2 = F3 plug in F2 and solve F3 = 28.8

If the tire has an acceleration

The acceleration means the Fnet is equal to ma = 20(2.6) =52N

Determine components of Fnet

Fnx = cos45(52) = 36.8

Fny = sin45(52) = 36.8

Sum of forces in x and y are equal to Fny and Fnx

F1x-F2x=Fnx cos(25)160 - cos(15)F2 = 36.8 solve for F2 = 112N

F1y -F2y -F3 = Fny sin(25)160 - sin(15)F2 - F3 = 36.8 plug in F2 and solve for F3 = 1.8N

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