Question 2.SP.9: A 200-kg cylinder is hung by means of two cables AB and AC t......
A 200-kg cylinder is hung by means of two cables AB and AC that are attached to the top of a vertical wall. A horizontal force P perpendicular to the wall holds the cylinder in the position shown. Determine the magnitude of P and the tension in each cable.
STRATEGY: Connection point A is acted upon by four forces, including the weight of the cylinder. You can use the given geometry to express the force components of the cables and then apply equilibrium conditions to calculate the tensions.
Final Answer
MODELING and ANALYSIS:
Free-Body Diagram. Choose point A as a free body; this point is subjected to four forces, three of which are of unknown magnitude. Introducing the unit vectors i, j, and k, resolve each force into rectangular components (Fig. 1):
\begin{aligned}\mathbf{P} & =P \mathbf{i} \\\mathbf{W} & =-m g \mathbf{j}=-(200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right) \mathbf{j}=-(1962 \mathrm{~N}) \mathbf{j}\quad(1)\end{aligned}For \mathbf{T}_{A B} \text { and } \mathbf{T}_{A C}, it is first necessary to determine the components and magnitudes of the vectors \overrightarrow{A B} \text { and } \overrightarrow{A C}. Denoting the unit vector along AB by \boldsymbol{\lambda}_{A B} \text {, you can write } \mathbf{T}_{A B} as
\begin{aligned}\overrightarrow{A B} & =-(1.2 \mathrm{~m}) \mathbf{i}+(10 \mathrm{~m}) \mathbf{j}+(8 \mathrm{~m}) \mathbf{k} \quad A B=12.862 \mathrm{~m} \\\boldsymbol{\lambda}_{A B} & =\frac{\overrightarrow{A B}}{12.862 \mathrm{~m}}=-0.09330 \mathbf{i}+0.7775 \mathbf{j}+0.6220 \mathbf{k} \\\mathbf{T}_{A B} & =T_{A B} \boldsymbol{\lambda}_{A B}=-0.09330 T_{A B} \mathbf{i}+0.7775 T_{A B} \mathbf{j}+0.6220 T_{A B} \mathbf{k}\quad (2)\end{aligned}Similarly, denoting the unit vector along AC by \boldsymbol{\lambda}_{A C} \text {, you have for } \mathbf{T}_{A C}
\begin{array}{l}\overrightarrow{A C}=-(1.2 \mathrm{~m}) \mathbf{i}+(10 \mathrm{~m}) \mathbf{j}-(10 \mathrm{~m}) \mathbf{k} \quad A C=14.193 \mathrm{~m} \\\boldsymbol{\lambda}_{A C}=\frac{\overrightarrow{A C}}{14.193 \mathrm{~m}}=-0.08455 \mathbf{i}+0.7046 \mathbf{j}-0.7046 \mathbf{k} \\\Gamma_{A C}=T_{A C} \boldsymbol{\lambda}_{A C}=-0.08455 T_{A C} \mathbf{i}+0.7046 T_{A C} \mathbf{j}-0.7046 T_{A C} \mathbf{k}\quad (3)\end{array}Equilibrium Condition. Since A is in equilibrium, you must have
\Sigma \mathbf{F}=0: \quad \mathbf{T}_{A B}+\mathbf{T}_{A C}+\mathbf{P}+\mathbf{W}=0or substituting from Eqs. (1), (2), and (3) for the forces and factoring i, j, and k, you have
\begin{array}{c}\left(-0.09330 T_{A B}-0.08455 T_{A C}+P\right) \mathbf{i} \\+\left(0.7775 T_{A B}+0.7046 T_{A C}-1962 \mathrm{~N}\right) \mathbf{j} \\+\left(0.6220 T_{A B}-0.7046 T_{A C}\right) \mathbf{k}=0\end{array}Setting the coefficients of i, j, and k equal to zero, you can write three scalar equations, which express that the sums of the x, y, and z components of the forces are respectively equal to zero.
\begin{array}{ll}\left(\Sigma F_x=0:\right) & -0.09330 T_{A B}-0.08455 T_{A C}+P=0 \\\left(\Sigma F_y=0:\right) & +0.7775 T_{A B}+0.7046 T_{A C}-1962 \mathrm{~N}=0 \\\left(\Sigma F_z=0:\right) & +0.6220 T_{A B}-0.7046 T_{A C}=0\end{array}Solving these equations, you obtain
P=235 \mathrm{~N} \quad T_{A B}=1402 \mathrm{~N} \quad T_{A C}=1238 \mathrm{~N}REFLECT and THINK: The solution of the three unknown forces yielded positive results, which is completely consistent with the physical situation of this problem. Conversely, if one of the cable force results had been negative, thereby reflecting compression instead of tension, you should recognize that the solution is in error.
