The sled in the figure below is stuck in the snow. A child pulls onthe rope and

Question

The sled in the figure below is stuck in the snow. A child pulls onthe rope and finds that the sled just barely begins to move when hepulls with a force of 24 N, with therope at an angle of 28 withrespect to the horizontal. (a) Sketch the sled and all the forces actingon it. Also choose a coordinate system. (Do this on paper. Yourinstructor may ask you to turn in this work.) (b) Determine the components of all the tension forces on the sledalong the coordinate axes. Tx = 1 N Ty = 2 N The weight will be in the 3 direction. The normal force N will be in the 4 direction. The friction force will be in the 5 direction. (c) Write the conditions for static equilibrium along your twocoordinate directions. (Use the following variables as necessary:T_x, T_y, m, g, F_friction, and N for the normal force.) Sigma Fx = 6 = 0 Sigma Fy = 7 = 0 (d) If the sled has a mass of 15 kg,what is the coefficient of friction between the sled and thesnow? 8 (e) Is this the coefficient of staticfriction or the coefficient of kinetic friction? coefficient of 9 friction

Explanation / Answer

T=24 N Tx =24*cos28=21.19 N Ty=24*sin28=11.27 N the weight is in -y direction the normal force N is in +y direction the frictional force is in -x direction Fx =Tx-Ffriction=0 or Fx=T cos28-Ffriction=0 N=T*sin28+m*g Fy=N-m*g=0 or Fy=N-m*g=0 when the static frictional force reaches it's maximum value thesled just begins to move m=15 T=24 Tcos28=24*cos28=21.19 g=9.8 =co-efficient of friction frictional force=*N frictional force=*N=21.19 (since the sled just begins tomove) 21.19=(Tsin28+m*g) 21.19=*(24*sin28+15*9.8) *158.27=21.19 =0.134 this is co-efficient of static friction

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