Explore BrainMass

Explore BrainMass

    A truck is to be driven upward on a plane. Find the maximum plane angles P, Q, S

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A truck whose weight is W, has a wheelbase (distance between front and back wheels) B= 3.2 meters. On level ground the front wheels carry 65% of its weight. The center of gravity (c.g.) of the truck is distance d= .85 m above the ground and unknown horizontal distance x, behind the front wheels' contact with the ground. The coefficient of friction is f= .70 between tires and plane.

    SEE ATTACHMENT #1 for a diagram with parameters.

    PART a. Find the distance x, from the front wheel to the c.g.
    PART b. Find the maximum slope angle P, for which the truck with only front wheel drive can move up the plane.
    PART c. Find the maximum slope angle Q, for which the truck with only rear wheel drive can move up the plane.
    PART d. Find the maximum slope angle S, for which the truck with four wheel drive can move up the plane.

    © BrainMass Inc. brainmass.com March 4, 2021, 5:42 pm ad1c9bdddf
    https://brainmass.com/physics/torques/find-maximum-plane-angles-6704

    Attachments

    Solution Preview

    PART a.
    We are given that on level ground the normal force on the front tires is 65% of the total weight. Therefore we can write:
    (1) NF= .65 W and
    (2) NR= .35 W
    To write an equation containing the wanted distance x, we can apply 'net torque=0' about any axis. Since x is the horizontal distance behind the front wheels' contact point with the ground we choose that point for an axis and express 'total ccl= total cl', using (2) for NR, to write:
    (3) (.35 W) (B) =(W) (x) Canceling W and substituting knowns gives:
    (4) x= .35 B = 1.12 m (The location of the c.g. does not change when the truck is on an angled plane.)

    PART b. ...

    Solution Summary

    With careful, step by step calculations, the solutions to the problem are clearly shown.

    $2.19

    ADVERTISEMENT