(6%) Problem 6: Two children pull a third child backward on a snow saucer sled e
ID: 1533210 • Letter: #
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(6%) Problem 6: Two children pull a third child backward on a snow saucer sled exerting forces F 7.5 and F 8.5 as shown in the figure. Note that the direction of the friction force f 5.4 Nis unspecified; it will be opposite in direction to the 45 sum of the other two forces. Free-body diagram Otheexpertta.com A 50% Part (a) Find the magnitude of the acceleration of the 45 kg sled and child system, in meters per second squared. Grade Summary Deductions Potential 100% tan() 7 8 9 sin cos Submissions cotano asin (O acoso 4 5 6 Attempts remaining: 10 4% per attempt atan( acotan sinh( 1 2 3 detailed view cosh() tanh cotanh0 Degrees O Radians Submit Hint I give up! Hints 30% deduction per hint. Hints remaining: 2 Feedback 4% deduction per feedback. 50% Part (b) Assuming the sled starts at rest, find the direction of the sled and child system in degrees north of east.Explanation / Answer
let horizontal be along +ve x axis and perpendicular to it is along +ve y axis.
let i and j are unit vectors along +ve x and +ve y axis repsectively.
then force F1=7.5*(cos(45) i+ sin(45) j)
=5.3033 i + 5.3033 j N
force F2=8.5*(cos(30) i-sin(30) j)
=7.3612 i - 4.25 j
so resultant force=F1+F2=12.665 i +1.0533 j N
angle with +ve x axis=arctan(1.0533/12.665)=4.7541 degrees
then friction force will be making an angle 4.7541 degrees below -ve x axis.
friction force in vector form=f=5.4*(-cos(4.7541) i -sin(4.7541) j)=-5.38142 i -0.44755 j N
so total force on the sled and child system
=F1+F2+f
==12.665 i +1.0533 j -5.38142 i -0.44755 j
=7.28358 i + 0.60575 j
then magnitude of net force=sqrt(7.28358^2+0.60575^2)=7.3087 N
magnitude of acceleration=force/mass
=7.3087/45=0.16242 m/s^2
part b:
angle with +ve x axis=arctan(0.60575/7.28358)=4.7541 degrees north of east
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