Respuesta :
Answer:
(a) reaction at each front wheel is 5272N (upward)
(b) force between boulder and pallet is 4124N (compression)
Explanation:
Acceleration of the truck [tex]a_{t[/tex] = 1 m/[tex]s^{2}[/tex] (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,
[tex]a_{A}[/tex] = 0.5 m/[tex]s^{2}[/tex] (upward) , [tex]a_{B}[/tex] = 0.5 m/[tex]s^{2}[/tex] (upward)
Let T be tension in the cable
pallet and boulder: ∑fy = ∑(fy)eff = 2T- ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])g = ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])[tex]a_{B}[/tex]
= 2T- (400 + 50)*(9.81 m/[tex]s^{2}[/tex]) = (400 + 50)*(0.5 m/[tex]s^{2}[/tex])
T = 2320N
Truck: [tex]M_{R}[/tex] = ∑([tex]M_{R}[/tex])eff: = [tex]-N_{f}[/tex] (3.4m) + [tex]m_{T}[/tex] (2.0m) - T (0.6m)= [tex]m_{T} a_{T}[/tex] (1.0m)
Nf = (2.0m)(2000 kg)(9.81 m/[tex]s^{2}[/tex] )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/[tex]s^{2}[/tex]) = 11541.2N - 409.4N - 588.2N = 10544N
∑fy (upward) = ∑(fy)eff: [tex]N_{f}[/tex] + [tex]N_{R}[/tex] - [tex]m_{T}[/tex]g = 0
10544 + [tex]N_{R}[/tex] - (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 0
[tex]N_{R}[/tex] = 9076N
∑fx (to the left) = ∑(fx)eff: [tex]F_{R}[/tex] - T = [tex]m_{T} a_{T}[/tex]
[tex]F_{R}[/tex] = 2320N + (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 4320N
(a) reaction at each front wheel:
1/2 [tex]N_{f}[/tex] (upward): 1/2 (10544N) = 5272N (upward)
(b) force between boulder and pallet:
∑fy (upward) = ∑(fy)eff: [tex]N_{B}[/tex] + [tex]M_{B}[/tex]g - [tex]m_{B}[/tex][tex]a_{B}[/tex]
[tex]N_{B}[/tex] = (400kg)(9.81 m/[tex]s^{2}[/tex]) + (400kg)(0.5 m/[tex]s^{2}[/tex]) = 4124N (compression)
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:
[tex]a_{T} =a_{A} +a_{B}[/tex]
Where
aA = acceleration of pallet = ?
aB = acceleration of boulder = ?
aA = aB
aT = acceleration of truck = 1 m/s²
[tex]1=a_{A} +a_{A}\\a_{A}=a_{B}=0.5m/s^{2}[/tex]
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:
[tex]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[/tex]
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:
[tex]-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[/tex]
From diagram 3 and 4:
∑Fy = ∑(Fy)ef
[tex]N+N_{R} -m_{T} g=0\\10532.2+N_{R}-(2000*9.8)=0\\N_{R}=9067.8N[/tex]
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
[tex]N_{B} +m_{B} g=m_{B} a_{B} \\N_{B}-(400*9.8)=(400*0.5)\\N_{B}=4120N[/tex]
