Two children pull a third child backward on a snow saucer sled exerting forces F

Question

Two children pull a third child backward on a snow saucer sled exerting forces F_1 and F_2 as shown in the figure. Note that the direction of the friction force f is unspecified; it will be opposite in direction to the sum of the other two forces. Randomized variables F_1 = 7.5 N F_2 = 5.5 N f = 5.5 N Find the magnitude of the acceleration of the 44 kg sled and child system, in meters per second squared. Assuming the sled starts at rest, find the direction of the sled and child system in degrees north of cast.

Explanation / Answer

Force in x direction is 7.5 cos 45 + 5.5cos(30) = 10.066 N

Force in Y direction = 7.5sin(45) - 5.5(sin(30) = 2.5533 N

direction of this force = atan( 2.5533/10.066) =14.23 degrees.

So friction acts opposite to this direction in 14.23 degrees

Net force would be sqrt( 10.066 ^2 + 2.5533^2) - 5.5 = 4.88 N

So acceleration will 4.88 / 44 = 0.111m/s^2

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