A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel-drive truck is 1 m/s2 , determine (a) the reaction at each of the front wheels, (b) the force between the boulder and the pallet.

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck = 1 m/  (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

=  0.5 m/ (upward) , =  0.5 m/ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( + )g =  ( + )

                                  = 2T- (400 + 50)*(9.81 m/) = (400 + 50)*(0.5 m/)

                        T = 2320N

Truck:   = ∑()eff: = (3.4m) + (2.0m) - T (0.6m)= (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/)  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: + - g = 0

                                       10544 +  - (2000kg)(9.81 m/ ) = 0

                = 9076N

   ∑fx (to the left) = ∑(fx)eff:   - T =

                                      = 2320N + (2000kg)(9.81 m/ ) = 4320N

(a) reaction at each front wheel:

1/2 (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: + g -

            = (400kg)(9.81 m/) + (400kg)(0.5 m/) = 4124N (compression)

a) The reaction at each of the fron wheels is 5266.1 N

b) The force between the boulder and the pallet is 4120 N

Explanation:

The acceleration of truck is:

Where

aA = acceleration of pallet = ?

aB = acceleration of boulder = ?

aA = aB

aT = acceleration of truck = 1 m/s²

From diagram 1 and 2, the system of external forces is:

∑Fy = ∑(Fy)ef (eq.1)

From diagram 1:

∑Fy = 2T - g(mA + mB)

Where T = tension force

mA = mass of pallet = 50 kg

mB = mass of boulder = 400 kg

From diagram 2:

∑(Fy)ef = aB(mA + mB)

Substituting into equation 1:

From diagram 3 and 4, represents the system of external forces:

∑MR = ∑(MR)ef (eq. 2)

From diagram 3:

∑MR = -N(2 + 1.4) + mTg(2) - T(0.6)

Where

N = normal force

mT = mass of truck = 2000 kg

From diagram 4:

∑(MR)ef = mTaT

Substituting into equation 2:

From diagram 3 and 4:

∑Fy = ∑(Fy)ef

a) The reaction at each of the front wheels is:

Rf = N/2 = 10532.2/2 = 5266.1 N

b) From diagram 5 and 6:

∑Fy = ∑(Fy)ef

the bal will not change its velocity.