Question

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.

Answer 1

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

WL = 609.26 lb

Step-by-step 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:

• Take moments about point B front wheel-road contact to follow equilibrium conditions.

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

• The sum of forces along the inclined surface = 0

u's*Na - W*sin(10) - P = 0

0.65*Na - 8000*sin(10) - P = 0

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

• Solve Eq1 and Eq2 simultaneously:

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:

• The sum of forces normal the inclined surface = 0

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

• The sum of forces along the inclined surface = 0

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

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

• Solve Eq1 and Eq2 simultaneously:

WL = 609.26 lb

Nc = 600 lb