8/8/2019 Car Incline Solutions Assign 7

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ENGN1217: Solutions to Problems in Assignment 7

Problem 7.1

The car is a rigid structure that can be represented as a truss as shown in the figurebelow. An external force, W, is applied at the top and it is resting on pin joints. Thereaction forces on the wheels are Na and Nb respectively. The dimensions are as given in

the problem.

FBD:

EME:

Fx = 0 (1)Fy = Na + Nb W = 0 (2)

MW = Na a + Nb b = 0 (3)From equations of mechanical equilibrium one can derive the reaction forces:

Na =b

a + bW, a = ao cos h sin (4)

Nb =a

a + bW, b = bo cos + h sin , (5)

In eqns (4) and (5) ao and bo are the values of a and b when = 0, and h is the heightof the centre of mass when = 0. There is an angle at which Na = Nb, and when = tan1 (ao/h) the centre of mass is directly above the rear wheel; Na = W and Nb = 0.

Resistance to free rolling is assumed to be zero. Resolving the force W into componentsparallel and perpendicular to the incline gives the force, Fr = W sin which is trying toroll the car down the incline (to reduce its potential energy).

The engine applies a torque to the wheels to propel the car up the incline at a constantspeed. The torque is applied either to the (i) rear, (ii) front, or (iii) rear and front wheelstogether. The result of the torque is a frictional force, Fa or Fb, or both.

Case 1: rear wheel drive, Fa Fr. Fb = 0

(Na cos ) W sin (6)

Substitution of relations (4) and (5) gives:

(bo cos + h sin ) (ao + bo)sin (7)

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For which the final solution is:

tan1

boao + bo h

= 10.2 (8)

Case 2: front wheel drive, Fb

Fr. Fa = 0

(Nb cos ) W sin (9)

Substitution of relations (4) and (5) gives:

tan1

aoao + bo + h

= 17.2 (10)

Case 3: 4 wheel drive, Fa + Fb Fr.

(Na + Nb)cos W sin (11)

Substitution of relations (4) and (5) gives:

tan1 [] = 26.6 (12)

Notice that solutions (8), (10) and (12), give:

= 0 for = 0 for all three cases.

= 45 for = 1 in case 3.

from (8) and (10) it can be shown that at = tan1 [(ao bo)/2h] = 44.2 front

and rear wheels will have equal contribution to drive.

Also, notice that in most multi-storey car parks the ramps are close to 20 and all carscan drive up without any problems. This means that the coefficient of friction betweenthe tyres and concrete floor is closer to 1.0 than 0.5.

An alternative solution is to assume a frame of reference with x-axis parallel and y-axisperpendicular to the incline. Then for the rear wheel drive we have:EME:

Fx = Fa W sin = 0 (13)Fy = Na + Nb W cos = 0 (14)

MW = Na ao + Nb bo = 0 (15)

In these equations Na and Nb act perpendicular to the incline (parallel to the y-axis).Naturally, the final solutions will be the same.

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Problem 7.2

The FBD (not shown) includes applied forces, WL and WP, weight of ladder and person,respectively, and reaction forces, Na from the floor, and Nb from the wall. The coefficientof friction with the floor is a, and the coefficient of friction with the wall is b.

Let be the y-axis, and be the x-axis. The angle is between the wall the ladder. The

length of the ladder is L. WL is at mid-point and WP is at the top of the ladder.EME:

Fx = Fa Nb = 0, where Fa = Na a (16)Fy = Na WL WP + Fb = 0, Fb = Nb b (17)

Mb = Fa L cos Na L sin + WL L/2sin = 0 (18)

From equation (18) we have:

tan = Fa

Na WL/2

(19)

Substitutions of relations (16) and (17) into (18) gives:

= tan1

a

1 12

1+ab1+WP/WL

(20)

Substitution of the values into (20) gives:

= tan1 [0.545] = 28.6 (21)

Note

The above solutions demonstrate the advantage of deriving the equations symbolically,rather than numerically. Consider equation (21) from which one can infer that:

1. the angle of the ladder depends primarily on the coefficient of friction with the floor.

2. the coefficient of friction with the wall is of secondary importance. Even if it is zero,it makes only a small difference to the angle (26.6).

3. The weight of the person plays a significant role. For example, ifWP = 0, the anglecan be increased to approx. 48.

None of these observations can be made if solutions are derived by numerical substitutionsmade too early.

Zbigniew H. StachurskiCollege of Engineering & Computer Science

Australian National University, Canberra, ACT 0200

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