q4Stx.theexpertta.com/Common/TakeTutorialAssignment.aspx (7%) Problem 14: Two ch
ID: 1577276 • Letter: Q
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q4Stx.theexpertta.com/Common/TakeTutorialAssignment.aspx (7%) Problem 14: Two children pull a third child backward on a snow saucer sled exerting forces F1 13.5 and F2- 6.5 as shown in the figure. Note that the direction of the friction force f- 5.8 N is unspecified, it will be opposite in direction to the sum of the other two forces. Free-body dagram Otheexpertta.com 50% Part (a) Find the magnitude of the acceleration of the 18 kg sled and child system, in meters per second squared. 50% Part (b) Assuming the sled starts at rest, find the direction of the sled and child system in degrees north of east. Grade Summary Deductions 096 Potential 100% sinO cotan0asin acos0 sinh0 ncotanh0 Submissions Attempts remaining: S 0% per attempt) detailed view atanO coshO Degrees Radians Submit Hint I give up Feedback: deduction per feedback Hints: deduction per hint Hints remaining: -- All content 2013 Expert TA, LLCExplanation / Answer
Using unit vectors, the resultant of the two forces is
R = (13.5N*cos45º + 6.5N*cos-30º) i + (13.5N*sin45º + 6.5N*sin-30º) j
R = 19.09N i + 3.20N j
mag |R| = (19.09² + 3.20²) N = 19.35 N
and this is opposed by a friction force of 5.8N, so the net force
Fnet = R - f = 13.55 N
The resulting acceleration is
a = Fnet / m = 13.55N / 48kg = 0.282 m/s²
Bonus: the actual direction of motion is
= arctan(3.07/14) = 9.51º above the horizontal line
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