Two children pull a third child on a snow saucer sled exerting forces
F1 and F2 as shown from above in Figure 4.36. Find the acceleration
of the 49.00-kg sled and child system. Note that the direction of the
frictional force is unspecified; it will be in the opposite direction of the sum
of F1=10N and F2=8 N
F1 have an angle 45 degree with horizontal line and F2 has an angle 30 degree below the horizontal line at fourth quadrant
also there is another force directed to the left on x axises = 7.5 N

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  1. F = 10N[45o] + 8N[330o] -7.5N

    X = 10*cos45 + 8*cos330 - 7.5 = 6.50 N.
    Y = 10*sin45 + 8*sin330 = 3.07 N.

    Tan A = Y/X = 3.07/6.50 = 0.47247
    A = 25.29o

    Fr = X/cosA = 6.50/cos25.29=7.19N[25.29]
    = Resultant force.

    a = Fr/m = 7.19[25.29o]/49=0.147[25.29]

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