Chapter 9: Q8P (page 230)
(II) The two trees in Fig. 9–51 are 6.6 m apart. A backpacker is trying to lift his pack out of the reach of bears. Calculate the magnitude of the force\(\vec F\)that he must exert downward to hold a 19-kg backpack so that the rope sags at its midpoint by (a) 1.5 m, (b) 0.15 m.
Short Answer
The magnitudes of forces for parts (a) and (b) are:
(a) \({F_{\rm{T}}} = 225.3\;{\rm{N}}\)
(b) \({F_{\rm{T}}} = 2076.2\;{\rm{N}}\)
Step by step solution
Given data
The distance between the two trees is\(L = 6.6\,{\rm{m}}\).
The mass of the backpack is\(m = 19\;{\rm{kg}}\).
The magnitude of the force is\(F\).
Understanding the direction made by the force
In this question, the backpack is midway between the two trees, and the angles in the diagram will be equivalent.
To find the angle of the force, use the distance between the two trees and the quantity of sag at the midpoint.
Free body diagram and calculation of magnitude and direction of force
The following is the free body diagram.
When the sag in the rope at the midpoint is \(y = 1.5\;{\rm{m}}\), the relation to calculate the angle of force can be written as:
\(\begin{array}{c}\tan \theta = \frac{y}{{\left( {\frac{L}{2}} \right)}}\\\tan \theta = \left[ {\frac{{1.5\,{\rm{m}}}}{{\left( {\frac{{6.6\,{\rm{m}}}}{2}} \right)}}} \right]\\\theta = 24.4^\circ \end{array}\)
The relation of the forces in the y-direction can be written as:
\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\\left[ {2{F_{\rm{T}}}\sin \theta - mg} \right] = 0\\{F_{\rm{T}}} = \frac{{mg}}{{2\sin \theta }}\end{array}\)
Here, \(g\) is the gravitational acceleration.
On plugging the values in the above relation, you get:
\(\begin{array}{l}{F_{\rm{T}}} = \left( {\frac{{19\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{2\left( {\sin 24.4^\circ } \right)}}} \right)\\{F_{\rm{T}}} = 225.3\;{\rm{N}}\end{array}\)
Thus, the magnitude of the force is \({F_{\rm{T}}} = 225.3\;{\rm{N}}\).
Calculation of magnitude and direction of force
When the sag in the rope at the midpoint is \(y = 0.15\;{\rm{m}}\), the relation to calculate the angle of force can be written as:
\(\begin{array}{c}\tan \theta ' = \frac{y}{{\left( {\frac{L}{2}} \right)}}\\\tan \theta ' = \left[ {\frac{{0.15\,{\rm{m}}}}{{\left( {\frac{{6.6\,{\rm{m}}}}{2}} \right)}}} \right]\\\theta ' = 2.57^\circ \end{array}\)
The relation of the forces in the y-direction can be written as:
\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\\left[ {2{F_{\rm{T}}}\sin \theta ' - mg} \right] = 0\\{F_{\rm{T}}} = \frac{{mg}}{{2\sin \theta '}}\end{array}\)
On plugging the values in the above relation, you get:
\(\begin{array}{l}{F_{\rm{T}}} = \left( {\frac{{19\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{2\left( {\sin 2.57^\circ } \right)}}} \right)\\{F_{\rm{T}}} = 2076.2\;{\rm{N}}\end{array}\)
Thus, the magnitude of the force is \({F_{\rm{T}}} = 2076.2\;{\rm{N}}\).
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