Chapter 9: Q62GP (page 230)
Question:The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over (Fig. 9–81)?
Short Answer
The slope at which the truck can be parked without tipping over is \(\theta = 28.6^\circ \).
Step by step solution
Given data
Given data
The height of the tree,\(h = 4\;{\rm{m}}\).
The width of the tree,\(b = 2.4\;{\rm{m}}\).
The center of gravity, \(CG = 2.2\;{\rm{m}}\).
Understanding the center of gravity
In this problem, the truck will not tip if the vertical line down from the center of gravity is among the wheels.
Also, the equilibrium will be unstable as the vertical line is at the wheel.
Free body diagram of the truck and calculation of the slope on which the truck can be parked
The following is the free body diagram.
The relation to calculate the slope angle can be written as follows:
\(\tan \theta = \frac{x}{{CG}}\)
Here, \(x\) is the horizontal distance whose value is half of the base.
Plugging in the values in the above relation:
\(\begin{array}{c}\tan \theta = \left( {\frac{{\left( {\frac{b}{2}} \right)}}{{CG}}} \right)\\\tan \theta = \left( {\frac{{\left( {\frac{{2.4\;{\rm{m}}}}{2}} \right)}}{{2.2\;{\rm{m}}}}} \right)\\\theta = 28.6^\circ \end{array}\)
Thus, \(\theta = 28.6^\circ \)is the required angle.
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