Problem

A crane consists of a quarter-circular bar that lies in a plane parallel to the xz plane...

A crane consists of a quarter-circular bar that lies in a plane parallel to the xz plane with a low-friction collar at point B that may slide on the bar. The forces supported by cables AB and BC are 30 lb and 1500 lb, respectively, and the coordinates of point A are given in the figure. For the value of angle α given below, determine expressions for the forces the two cables apply to the collar at point B.

Step-by-Step Solution

Solution 1

Draw the forces supported by cables AB and BC as follow.

Picture 15

For \(\boldsymbol{\alpha}=\mathbf{0}^{\circ}\) :

Determine the position vector between \(A\) and \(B\), using the relation,

$$ \vec{r}_{i s}=\left(\left\{x_{1 s}-x_{11}\right) \hat{i}+\left(y_{i s}-y_{i}\right) \hat{j}+\left(z_{n}-z_{i l}\right) \hat{k}\right) \mathrm{ft} $$

Substitute the coordinates for point \(A\), which are \((8,4.5,2) \mathrm{ft}\), and the coordinates for point \(B\), which are \((0,10,5) \mathrm{ft}\).

\(=((0-8) \hat{i}+(10-4.5) \hat{j}+(5-2) \hat{k}) \mathrm{fl}\)

\(=((-8) \hat{i}+5,5 \hat{j}+3 \hat{k})+\mathrm{t}\)

$$ \left|\dot{r}_{A A}\right|=\sqrt{r_{\cdot}^{2}+r_{1}^{2}+r_{-}^{2}} $$

Here, \(\vec{r}_{i s}=\left(r_{1} \hat{i}+r \hat{j}-r_{i} \hat{k}\right)\).

Substitute \((-8 \mathrm{ft})\) for \(\boldsymbol{r}_{x},(55 \mathrm{ft})\) for \(r_{y}\), and \((3 \mathrm{ft})\) for \(r_{z}\)

$$ \begin{aligned} \left|r_{A s}\right| &=\sqrt{(-8 \mathrm{ft})^{2}+(5.5 \mathrm{ft})^{2}+(3 \mathrm{ft})^{2}} \\ &=10.16 \mathrm{ft} \end{aligned} $$

Calculate the cable force exerted on A using the relation,

Substitute the position vector and the magnitude and solve for .

The force that cable AB exerts on the support B is and according to Newton’s third law, it is of equal magnitude and opposite direction to the force the cable exerts on A. Therefore, .

For :

Determine the coordinates of B, where .

Picture 17

Find using the bar radius (5 ft) and the angle, .

Find using the bar radius (5 ft), the angle, .

Therefore, the coordinates of B are (5,10,0)ft.

Determine the position vector between A and B, using the relation,

Substitute the coordinates for point A, which are (8,4.5,2)ft, and the coordinates for point B, which are (5,10,0)ft.

Use to calculate the magnitude of the vector, , using the relation,

Here, .

Substitute , , and .

Calculate the cable force exerted on A using the relation,

Substitute the position vector and the magnitude and solve for .

The force that cable AB exerts on the support B is and according to Newton’s third law, it is of equal magnitude and opposite direction to the force the cable exerts on A. Therefore, .

Therefore, , between , is

The vertically hanging cable, BC, supports 1500 lbs.

Determine the force, .

Therefore, the force exerted by the cable BC on point B is , no matter what the angle of .

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