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.

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.

Accepted Answer

Kinematics: Acceleration of truck: [latex]{\bf a}_{T}=1\mathrm{\ m}/\mathrm{ s}^{2}\leftarrow[/latex].

When the truck moves 1 m to the left, the boulder B and pallet A are raised 0.5 m.

Then, [latex]{\bf a}_{A}=0.5~\mathrm{m/s}^{2}\uparrow\quad\quad\quad\quad\mathbf{a}_{B}=0.5~\mathrm{m/s}^{2}\uparrow[/latex]

Kinetics: Let T be the tension in the cable.

Pallet and boulder:

[latex]\begin{aligned}& +\uparrow\Sigma F_{y}=\Sigma(F_{y})_{\mathrm{eff}};\quad 2T-(m_{A}+m_{B})_{g}=(m_{A}+m_{B})a_{B}\& \quad 2T-(450\mathrm{~kg})(9.81\mathrm{~m/s^{2}})=(450\mathrm{~kg})(0.5\ \mathrm{m/s^{2}})\& \quad T=2320\mathrm{\ N}\end{aligned}[/latex]

Truck:

[latex]\begin{aligned}& +\circlearrowleft M_{R}=\Sigma(M_{R})_{\mathrm{eff}};~~~-N_{F}(3.4~\mathrm{m})+m_{T}g(2.0~\mathrm{m})-T(0.6~\mathrm{m})=m_{T}a_{T}(1.0~\mathrm{m})\& \quad N_{F}=\frac{(2.0~\mathrm{m})(2000~\mathrm{kg})(9.81~\mathrm{m}/\mathrm{s}^{2})}{3.4~\mathrm{m}}-\frac{(0.6~\mathrm{m})(2320~\mathrm{N})}{3.4~\mathrm{m}}+\frac{(1.0~\mathrm{m})(2000~\mathrm{kg})(1.0~\mathrm{m}/\mathrm{sg})}{3.4~\mathrm{m}}\& \quad\quad ={11541.2\ \mathrm{N}-{409.4\ \mathrm{N}-588.2\ \mathrm{N}}}\& \quad\quad ={10544\,\mathrm{N}}\& +\uparrow\Sigma F_{y}=\Sigma(F_{y})_{\mathrm{eff}}\,;~~~~N_{F}+N_{R}-m_{T}g=0\& \quad\quad\quad\quad\quad\quad\quad\quad 10544\ {\mathrm{N}}+{N}_{R}-(2000\ {\mathrm{kg}})(9.81\ {\mathrm{m/s}}^{2})=0\& \quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad N_{R}=9076\ \mathrm{N}\& \overset{+}{\longleftarrow } F_{x}=\Sigma(F_{x})_{\mathrm{eff}}\!:\quad F_{R}-T=m_{T}a_{T}\& \quad\quad\quad\quad\quad\quad\quad\quad F_{R}=2320\,\mathrm{N}+(2000\,\mathrm{kg})(1.0\,\mathrm{m}/\mathrm{s}^{2})\& \quad\quad\quad\quad\quad\quad\quad\quad\quad =4320\mathrm{\ N}\end{aligned}[/latex]

(a) Reaction at each front wheel:

[latex]\frac{1}{2}N_{F}\uparrow:[/latex]

[latex]5270\,\mathrm{N}\uparrow\blacktriangleleft[/latex]

Reaction at each rear wheel:

[latex]\frac{1}{2}F_{R}\leftarrow +\frac{1}{2}N_{R}\uparrow[/latex]

[latex]5030\,\mathrm{N}\ ⦩64.5^{\circ}[/latex]

(b) Force between boulder and pallet.

Boulder

[latex]\begin{aligned}& +\uparrow\Sigma F_{y}=\Sigma(F_{y})_{\mathrm{eff}}\,;\;\;\;\;\;N_{B}+M_{B}g-m_{B}a_{B}\& \quad\quad N_{B}=(400\ \mathrm{kg})(9.81\ \mathrm{m/s}^{2})+(400\ \mathrm{kg})(0.5\ \mathrm{m/s}^{2})\& \quad\quad\quad\quad =4124{\mathrm{\ N}} & 4120\mathrm{\ N} ext{(compression)}\blacktriangleleft\end{aligned}[/latex]

Question 16.6: A 2000-kg truck is being used to lift a 400-kg boulder B tha...

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