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

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

Step-by-step 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=(a_(B)(m_(A) +m_(B)) +g(m_(A) +m_(B) ) )/(2) =(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=((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

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

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

Step-by-step 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)