Determine the greatest weight of the log that can be pushed up the incline. The coefficient of static friction between the log and the ground is μs = 0.50, and between the rear wheels of the tractor and the ground μ′s = 0.65. The front wheels are free to roll. Assume the engine can develop enough torque to cause the rear wheels to slip.

The greatest weight of the log that can be pushed up the surface is:

                          WL = 609.26 lb

Explanation:

Given:-

- The weight of the tractor, W = 8000 lb

- Coefficient of static friction between log and ground, us = 0.50

- Coefficient of static friction between wheel and ground, u's = 0.65

Solution:-

- Use the given figure, and make a free body diagram of the tractor marking 5 forces acting on the Tractor.

Force1 : The weight "W" f the tractor directed vertically down from its Center of gravity.

Force 2: The contact force (Na) at the rear wheel-road contact normal to inclined surface.

Force 3: The contact force (Nb) at the front wheel-road contact normal to inclined surface.

Force 4: The friction force (Fa) at the rear wheel-road contact parallel to inclined surface directed uphill ( against the tipping motion). = u's*Na

Force 5: The reaction force (P) exerted on the tractor's claw  

Note: The friction for for front wheel is not applicable under the assumption that front wheels are free to roll.

- See the attachment for the complete FBD for the tractor.

Tractor:

       W*cos(10)*(3) + W*sin(10)*(2.5) + P*(1.25) - Na*(10) = 0

       8000*cos(10)*(3) + 8000*sin(10)*(2.5) + P*(1.25) - Na*(10) = 0

       Na - P*0.125 = 2710.8  .... Eq 1

       u's*Na - W*sin(10) - P = 0

       0.65*Na - 8000*sin(10) - P = 0

       0.65*Na - P = 1389.2  ..... Eq2

        P = 405.79 lb

        Na = 2761.52 lb

- Use the given figure, and make a free body diagram of the log marking 4 forces acting on the log.

Force1 : The weight "WL" of the log directed vertically down from its Center of gravity.

Force 2: The contact force (Nc) at the log-road contact normal to inclined surface.

Force 3: The friction force (Fc) at the  log-road contact parallel to inclined surface directed downhill ( against the upward push motion). = us*Nc

Force 4: The reaction force (P) exerted on the log by the tractor claw  

- See the attachment for the complete FBD for the Log.

Log:

       Nc - WL*cos(10)  = 0  .... Eq 1

       405.79 - WL*sin(10) - 0.5 Nc = 0

      WL*sin(10) + 0.5Nc = 405.79  ..... Eq2

        WL = 609.26 lb

        Nc = 600 lb

1

explanation:

answer:

a

explanation: