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Mobile App Chapter 6: Q1P (page 138)
(I) Three forces are applied to a tree sapling, as shown in Fig. 9–46, to stabilize it. If\({\vec F_{\rm{A}}}=385\;{\rm{N}}\)and\({\vec F_{\rm{B}}} = 475\;{\rm{N}}\), find\({\vec F_{\rm{C}}}\)in magnitude and direction.
Short Answer
The magnitude and direction are \(528.3\;{\rm{N}}\) and \(119.8^\circ \).
Step by step solution
Given Data
The force on point A is\({\vec F_{\rm{A}}} = 385\;{\rm{N}}\).
The force on point B is\({\vec F_{\rm{B}}} = 475\;{\rm{N}}\).
The angle is\(\alpha = 105^\circ \).
Evaluate the net force on the tree
In this problem, three forces are applied on the tree. As the tree is not accelerating, the net force on it from all directions is zero.
Evaluate the forces in the x- and y-directions
The following is the free-body diagram of the tree.
Consider the horizontal direction as the x-axis and the vertical direction as the y-axis.
The relation for the forces in the x-direction can be written as shown below:
\(\begin{array}{c}\sum {F_{\rm{x}}} = {F_{\rm{A}}} + {F_{\rm{B}}}\cos \alpha + {F_{{\rm{Cx}}}}\\0 = {F_{\rm{A}}} + {F_{\rm{B}}}\cos \alpha + {F_{{\rm{Cx}}}}\\{F_{{\rm{Cx}}}} = - \left( {385\;{\rm{N}} + \left( {475\;{\rm{N}}} \right)\cos 105^\circ } \right)\\{F_{{\rm{Cx}}}} = - 262.06\;{\rm{N}}\end{array}\)
Here,\({F_{{\rm{Cx}}}}\)is the force at C in the x-direction.
The relation for forces in the y-direction can be written as shown below:
\(\begin{array}{c}\sum {F_{\rm{y}}} = {F_{\rm{B}}}\sin \alpha + {F_{{\rm{Cy}}}}\\0 = {F_{\rm{B}}}\sin \alpha + {F_{{\rm{Cy}}}}\\{F_{{\rm{Cy}}}} = - \left( {475\;{\rm{N}}\sin 105^\circ } \right)\\{F_{{\rm{Cy}}}} = - 458.8\;{\rm{N}}\end{array}\)
Here,\({F_{{\rm{Cy}}}}\)is the force at C in the y-direction.
Calculate the magnitude of force
The formula for the magnitude of force is
\({F_{\rm{C}}} = \sqrt {{{\left( {{F_{{\rm{Cx}}}}} \right)}^2} + {{\left( {{F_{{\rm{Cy}}}}} \right)}^2}} \).
Put the values in the above relation.
\(\begin{array}{l}{F_{\rm{C}}} = \sqrt {{{\left( { - 262.06\;{\rm{N}}} \right)}^2} + {{\left( { - 458.8\;{\rm{N}}} \right)}^2}} \\{F_{\rm{C}}} = 528.3\;{\rm{N}}\end{array}\)
Calculate the direction of force
The formula for the direction of force is\(\tan \theta = \left( {\frac{{{F_{{\rm{Cy}}}}}}{{{F_{{\rm{Cx}}}}}}} \right)\).
Put the values in the above relation.
\(\begin{array}{c}\tan \theta = \left( {\frac{{ - 458.8\;{\rm{N}}}}{{262.06\;{\rm{N}}}}} \right)\\\theta = 60.2^\circ \end{array}\)
The required direction of force at C is calculated below:
\(\begin{array}{l}\phi = 180^\circ - \theta \\\phi = 180^\circ - 60.2^\circ \\\phi = 119.8^\circ \end{array}\)
Thus, \({F_{\rm{C}}} = 528.3\;{\rm{N}}\) and \(\phi = 119.8^\circ \) are the required magnitude and direction of force.
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