Question

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.

Answer 1

Answer:

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:

$a_{T} =a_{A} +a_{B}$

Where

aA = acceleration of pallet = ?

aB = acceleration of boulder = ?

aA = aB

aT = acceleration of truck = 1 m/s²

$1=a_{A} +a_{A}\\a_{A}=a_{B}=0.5m/s^{2}$

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:

$2T-g(m_{A} +m_{B} )=a_{B} (m_{A} +m_{B} )\\T=\frac{a_{B}(m_{A} +m_{B}) +g(m_{A} +m_{B} ) }{2} =\frac{0.5(50+400)+9.8(50+400)}{2} =2317.5N$

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:

$-N(2+1.4)+m_{T} g(2)-T(0.6)=m_{T} a_{T} \\-N(3.4)+(2000*9.8*2)-(2317.5*0.6)=2000*1\\N=\frac{(2000*9.8*2)-(2317.5*0.6)-2000}{3.4} =10532.2N$

From diagram 3 and 4:

∑Fy = ∑(Fy)ef

$N+N_{R} -m_{T} g=0\\10532.2+N_{R}-(2000*9.8)=0\\N_{R}=9067.8N$

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

$N_{B} +m_{B} g=m_{B} a_{B} \\N_{B}-(400*9.8)=(400*0.5)\\N_{B}=4120N$

Answer 2

Answer:

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

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

Explanation:

Acceleration of the truck $a_{t$ = 1 m/$s^{2}$  (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,

$a_{A}$ =  0.5 m/$s^{2}$ (upward) , $a_{B}$ =  0.5 m/$s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ($m_{A}$ + $m_{B}$)g =  ($m_{A}$ + $m_{B}$)$a_{B}$

                                  = 2T- (400 + 50)*(9.81 m/$s^{2}$) = (400 + 50)*(0.5 m/$s^{2}$)

                        T = 2320N

Truck:  $M_{R}$ = ∑($M_{R}$)eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/$s^{2}$ )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/$s^{2}$)  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$g = 0

                                       10544 + $N_{R}$  - (2000kg)(9.81 m/$s^{2}$ ) = 0

                $N_{R}$ = 9076N

   ∑fx (to the left) = ∑(fx)eff:  $F_{R}$ - T = $m_{T} a_{T}$

                                      $F_{R}$ = 2320N + (2000kg)(9.81 m/$s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$g - $m_{B}$$a_{B}$

            $N_{B}$ = (400kg)(9.81 m/$s^{2}$) + (400kg)(0.5 m/$s^{2}$) = 4124N (compression)