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

Question

The sled in the figure below is stuck in the snow. A child pulls on the rope and finds that the sled just barely begins to move when he pulls with a force of 22 N, with the rope at an angle of 26 degree with respect to the horizontal. Sketch the sled and all the force acting on it. Also choose a coordinate system. (Do this on paper. Your instructor may ask you to turn in this work.) Determine the components of all the tension forces on the sled along the coordinate axes. Tx = N Ty = N The weight will be in the direction. The normal force N will be in the direction. The friction force will be in the direction Write the conditions for static edquilibrium along your two coordinates directions. (Use the following as necessary: Tx', Ty', m, g, Ffriction, and N to represents the magnitudes of these values.) Fx = = 0 Fy = 0 If the sled has a mass of 13 kg, what is the coefficient of friction between the sled and the snow? mu = Is this the coefficient of static friction or the coefficient of kinetic friction? coefficient of friction

Explanation / Answer

Tcos26= frictional force, Tsin26= normal force,, weight downward, normal upwards, friction opposite to motion, for horrrizontal equilibrium Tcos26= frictional force and vertical Tysin26= normal force, d) 22cos26 = mu*13*g, block is stationery so static friction

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