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Jobisdone [24]
4 years ago
10

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.

Physics
2 answers:
USPshnik [31]4 years ago
6 0

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

anzhelika [568]4 years ago
3 0

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)

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