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vladimir1956 [14]
3 years ago
13

A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a spee

d of 17 knots. (a) determine the magnitude and direction of the wind velocity as measured on the boat. (b) what is the component of the wind velocity in the direction parallel to the motion of the boat?
Physics
1 answer:
stellarik [79]3 years ago
5 0
(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
 
                   R = sqrt ((20 knots)² + (17 knots)²)
                    R = sqrt (400 + 289)
                      R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
                   cos θ = (20 knots)/(26.24 knots)
                        θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero. 
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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
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Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

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          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

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