Kim and Kent are on opposite sides of a canal, pulling a barge with tow ropes. Kim exerts a force of 50N at 20 degrees to the canal and Kent pulls at an angle of 15 degrees with just enough force so that the resultant force vector is directly along the canal. Find the number of Newtons with which Kent must pull and the magnitude of the resultant vector.

Draw a diagram. Let the barge be in the canal, with Kim on the south shore, and Kent on the north shore, both pulling eastward. Then we have

Kim: u = <50cos20,-50sin20>
u = <46.98,-17.10>

Kent: v = <kcos15,ksin15>
= <0.966k,0.259k>

You want the resultant u+v to point eastward, so

-17.10 + 0.259k = 0
k = 66.02

So, Kent pulls with a force of 66 N.

u+v = <46.98+63.78,0> = <110.76,0>

To solve this problem, we will break down the forces and use vector addition to find the resultant force.

Let's start by finding the x and y components of Kim's force:

Fx (Kim) = 50N * cos(20°)
Fy (Kim) = 50N * sin(20°)

Next, let's find the x and y components of Kent's force:

Fx (Kent) = F (Kent) * cos(15°)
Fy (Kent) = F (Kent) * sin(15°)

Since the resultant force vector is directly along the canal, the sum of the x components and y components must be zero. Therefore, we can set up the following equations:

Fx (Kim) + Fx (Kent) = 0
Fy (Kim) + Fy (Kent) = 0

Now, substitute the expressions for the x and y components:

50N * cos(20°) + F (Kent) * cos(15°) = 0
50N * sin(20°) + F (Kent) * sin(15°) = 0

Solve these equations to find the value of F (Kent). First, divide the second equation by the first equation:

(50N * sin(20°) + F (Kent) * sin(15°)) / (50N * cos(20°) + F (Kent) * cos(15°)) = 0

Then, cross-multiply:

50N * sin(20°) + F (Kent) * sin(15°) = 0

Rearrange the equation to isolate F (Kent):

F (Kent) * sin(15°) = -50N * sin(20°)

Divide both sides by sin(15°) to solve for F (Kent):

F (Kent) = (-50N * sin(20°)) / sin(15°)

Using a calculator, we find that F (Kent) ≈ -102.669N (rounded to three decimal places).

Note: The negative sign indicates that Kent's force is in the opposite direction to Kim's force.

To find the magnitude of the resultant vector, we can use the Pythagorean theorem:

Resultant = √((Fx(Kim) + Fx(Kent))^2 + (Fy(Kim) + Fy(Kent))^2)

Plug in the values:

Resultant = √((50N * cos(20°) + (-102.669N * cos(15°)))^2 + (50N * sin(20°) + (-102.669N * sin(15°)))^2)

Using a calculator, we find that the magnitude of the resultant vector ≈ 115.654N (rounded to three decimal places).

Therefore, Kent must pull with approximately 102.669 Newtons of force, and the magnitude of the resultant vector is approximately 115.654 Newtons.