C O wwww.webassi 6554939 s a top-down view of two children pulling a 11.2-kg sle

Question

C O wwww.webassi 6554939 s a top-down view of two children pulling a 11.2-kg sled along the snow. The first child exerts a force of F, 14 N at an angle 01-45counterclockwise from the positive x direction. The second child exerts a force of F2-6 N at an angle 02-30° clockwise from the positive x direction. (a) Find the magnitude and direction of the friction force acting on the sled if it moves with constant velocity magnitude 15.46 direction ·counterclockwise from the +x-axis (b) What is the coefficient of kinetic friction between the sled and the ground? 1408 (c) What is the magnitude of the acceleration of the sled if F, is doubled and Fz is halved in magnitude? m/s?

Explanation / Answer

F1 = 14 N, F2 = 6 N

Net force along x axis is, Fx = F1 cos 45 + F2 cos 30 = 14*0.707 + 6 * 0.866 = 9.898 + 5.196 = 15.094 N

Net force along y axis is, Fy = = F1 sin 45 + F2 sin 30 = 9.898 – 3 = 6.898 N

Magnitude of the net force due to application of the two forces is, F = Fx² + Fy² = 227.83 + 47.58 = 16.59 N

Since, the sled moves with the constant velocity Fnet = 0

i.e. F = fs

i.e. force of friction fs = 16.59 N (Answer)

And the direction in counterclockwise direction of x axis is given by,

Tan = Fy/Fx = 6.898/15.094 = 0.457

So, = tan-1(0.457) = 24.560 (Answer)

b) To get the value of coefficient of kinetic friction we use the force of friction from part a.

µ = fs/N = 16.59/mg = 16.59/(11.2*9.8) = 0.1511 (Answer)

c) Now, F1 is doubled while F2 is halved, then the net applied force F is given by,

F1 = 28 N, F2 = 3 N

Net force along x axis is, Fx = F1 cos 45 + F2 cos 30 = 28*0.707 + 3 * 0.866 = 19.796 + 2.598 = 22.394 N

Net force along y axis is, Fy = = F1 sin 45 + F2 sin 30 = 28 * 0.707 – 3 * 0.5 = 19.796 – 1.5 = 18.296 N

Magnitude of the net force is, F = Fx² + Fy² = 501.49 + 334.74 = 28.91 N

But, we know that, Fnet = F - fs

i.e. m*a = 28.91 - (0.1511 * 11.2 * 9.8) = 28.91 - 16.584 = 12.326

a = 12.326/11.2 = 1.1 m/s² (Answer)

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