Home Student: aals Class Management Help nit 4-Newton\'s Laws of Motion egin Dat

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Home Student: aals Class Management Help nit 4-Newton's Laws of Motion egin Date: 2/1/2018 10.00:00 AM-Due Date: 2/8/2018 11:00.00 PM End Date: 5/4/2018 11:00 (7%) Problem!4: Two children nall 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 isi specified; it will be opposite in direction to the sum of the other two forces. 45 30° Free-body diagram F, Otheexpert 50% Part ( Findthe magnitude of the acceleration of the 48 kg sled and child system, in meters per second squared. 50% Part b) Assuming the sled starts at ,est find the direction of the sled and child systemin degrees north of east. Grade Summ Deductions Potential sino I cosO | tan() | | (| | 7| 8| 9 cotanOasinOacosO atanO acotanOsinhO coshO tanhc Submissions Attempts remais (Ollo per attempt detailed vien 45 6 1 2 3 cotanhO Degrees Radians Submit Hint I give up

Explanation / Answer

At first we must find the resultant force of F1 and F2. To find it, we find x and y components of F1 and F2:

(Note: we call the resultant force of F1 and F2 as "F")

F1x = F1 cos(45) = 13.5 cos45= 9.546 N

F1y = F1 sin(45) = 13.5 sin(45) = 9.546 N

F2x = F2 cos(30) = 6.5 * cos(30) = 7.361 N

F2y = -F2 sin(30) = -6.5* sin(30) = - 3.42N

Now we can find x and y components of F:

Fx = F1x + F2x = 9.546+ 7.361 = 16.907 N

Fy = F1y + F2y = 9.546 + (-3.48 ) = 6.o66 N

Magnitude of the F:

F = sqrt(Fx^2 + Fy^2)

F = sqrt(16.902^2 + 6.356^2)

    = 18.057 N

Net force on the children will be:

F_net = F - (friction force) = F - f = 18.057 - 5.6 = 12.457 N

a = (F_net)/m

    = 12.066/48

    = 0.2594 m/s^2

direction = arctan( 6.356 / 16.907)

              = 20.06 degrees

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