A 325 Kg boat is sailing 15 degrees north of east at a speed of 2.00 ms/s. thirt

Question

A 325 Kg boat is sailing 15 degrees north of east at a speed of 2.00 ms/s. thirty seconds later, it is sailing 35 degrees north of east at a speed of 4.00 m/s. During this time tree forces act on the boat: a 31.0N force directed 15.0 degrees north of east (due to the auxiliary engine), a 23.0N force directed 15.0 degrees south of west( resistance due to the water), and Fw (due to the wind). Find the magnitude and direction of the force Fw. Express the direction as an angle with respect due to east.

Explanation / Answer

v1 = 2.0 *cos(15) + 2.0 * sin(15) m/s
v2 = 4.0 *cos(35) + 4.0 * sin(35) m/s

Force = m*a
Fnet = m*(v2-v1)/t


F1 = 31.0 * cos(15) i^ + 31.0 * sin(15) j^ N
F2 = - 23.0 * cos(15) i^ - 23.0 * sin(15) j^ N
Fw = x i^ + y j^ N

Taking the X component,
325*(4.0 *sin(35) -2.0 *sin(15))/30 = [31.0 * sin(15) - 23.0 * sin(15) + Fwy]
Fwx = 6.84 N

Taking the Y component,
325*(4.0 *cos(35) -2.0 *cos(15))/30 = [31.0 * cos(15) -  23.0 * cos(15) + Fwx]
Fwy = 17.2 N

Fw = sqrt(Fwx^2 + Fwy^2)
Fw = sqrt(6.84^2 + 17.2^2) N
Fw = 18.5 N

Direction = tan^-1(17.2/6.84)
Direction =68 o North of East

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