Question

a crate is supported by three cables as shown. determine the weight w of the crate, knowing that the tension in cable ad is 924 lb. the weight w of the crate is \boxed{} lb.

Explanation:

Step1: Find the length of AO

First, we need to find the length of \( AO \). From the diagram, we can see that the horizontal distance from \( A \) to the \( y - z \) plane is 45 in, and the distance from \( D \) to \( O \) in the \( y - z \) plane: let's consider the coordinates. Let's assume \( O \) is at the origin \((0,0,0)\). For point \( D \), the distance from \( O \) to \( D \) in the \( y - z \) plane: we have a right triangle with legs 18 in and 26 in? Wait, no, actually, the length of \( AD \) can be found using the distance formula in 3D. Wait, the vertical distance (along \( x \)-axis) from \( A \) to \( O \) is 45 in? Wait, no, looking at the diagram, the distance from \( A \) to \( O \) along the \( x \)-axis is 45 in, and the distance from \( D \) to \( O \) in the \( y - z \) plane: let's calculate the length of \( OD \). From the diagram, the segments are 18 in and 26 in? Wait, no, the length of \( AD \) is the hypotenuse of a right triangle with one leg 45 in (along \( x \)-axis) and the other leg being the distance from \( D \) to \( O \) in the \( y - z \) plane. Wait, let's calculate the length of \( OD \). The coordinates of \( D \) relative to \( O \): in the \( y - z \) plane, the horizontal (y) and vertical (z) distances? Wait, the length of \( OD \) can be calculated using the Pythagorean theorem for the two segments: 18 in and 26 in? Wait, no, 18 in and 26 in: \( \sqrt{18^2 + 26^2} \)? Wait, no, maybe the length of \( AD \) is \( \sqrt{45^2 + OD^2} \), where \( OD \) is \( \sqrt{18^2 + 26^2} \)? Wait, no, let's check the diagram again. The length of \( AD \) is given by the distance between \( A \) and \( D \). The coordinates: let's set \( O \) at \((0,0,0)\), \( A \) at \((45,0,0)\) (since the distance from \( A \) to \( O \) along \( x \)-axis is 45 in), and \( D \) at \((0, - 18,26)\) (assuming the directions). Then the distance \( AD \) is \( \sqrt{(45 - 0)^2+(0 + 18)^2+(0 - 26)^2}=\sqrt{45^2 + 18^2+26^2} \). Let's calculate that: \( 45^2 = 2025 \), \( 18^2 = 324 \), \( 26^2 = 676 \). Sum: \( 2025+324 + 676=3025 \). So \( AD=\sqrt{3025}=55 \) in.

Step2: Find the vertical component of tension in AD

The tension in cable \( AD \) is \( T_{AD} = 924 \) lb. The vertical component (along \( x \)-axis? Wait, no, the weight \( W \) is vertical, so we need the component of \( T_{AD} \) along the vertical direction (which is opposite to the weight? Wait, no, the crate is in equilibrium, so the sum of the vertical components of the tensions in the cables should equal the weight \( W \). Wait, actually, the vertical direction here is along the line from \( A \) to the crate, which is vertical (since the crate is hanging, so the weight is vertical downward, and the cables' vertical components are upward). Wait, the length of \( AD \) is 55 in, and the vertical distance (the component along the line from \( A \) to the crate, which is the same as the \( x \)-axis distance? Wait, no, the vertical component (the component that balances the weight) is the component of \( T_{AD} \) along the direction of the weight, which is the direction from \( A \) to the crate, i.e., the vertical direction. Wait, the length of \( AD \) is 55 in, and the vertical segment (the distance from \( A \) to the crate's attachment point) is 45 in? Wait, no, the vertical component (the component that contributes to balancing the weight) is the ratio of the vertical distance (45 in) to the length of \( AD \) (55 in) times the tension \( T_{AD} \). So the vertical component \( F_{ADz} = T_{AD}\times\frac{45}{55} \).

Step3: Calculate the vertical component

\( F_{ADz}=924\times\frac{45}{55} \). Simplify \( \frac{45}{55}=\frac{9}{11} \). So \( 924\times\frac{9}{11}=84\times9 = 756 \) lb. Wait, but is this the only cable? Wait, no, the problem says three cables, but we are given tension in \( AD \), and we need to find the weight. Wait, maybe the system is in equilibrium, so the sum of the vertical components of all cables equals \( W \). But maybe in this case, the vertical component of \( AD \) is equal to \( W \)? Wait, no, maybe the diagram is such that the vertical direction is along the \( x \)-axis, and the other cables have horizontal components, but the weight is balanced by the vertical component of \( AD \). Wait, maybe I made a mistake. Wait, let's re-examine the coordinates. Let's define the coordinate system with \( O \) at the origin, \( x \)-axis along \( AO \) (45 in), \( y \)-axis and \( z \)-axis in the plane perpendicular to \( AO \). Then the vector \( \overrightarrow{AD} \) goes from \( A(45,0,0) \) to \( D(0, - 18,26) \), so the vector \( \overrightarrow{AD}=(- 45,-18,26) \). The length of \( \overrightarrow{AD} \) is \( \sqrt{(-45)^2+(-18)^2 + 26^2}=\sqrt{2025 + 324+676}=\sqrt{3025}=55 \) in, as before. The unit vector in the direction of \( AD \) is \( \frac{(-45,-18,26)}{55} \). The tension vector \( \mathbf{T}_{AD}=T_{AD}\times\frac{(-45,-18,26)}{55} \). The weight \( \mathbf{W}=(0,0,-W) \) (assuming downward is negative z, but maybe upward is positive z). Wait, the equilibrium condition is \( \sum\mathbf{F}=0 \), so the sum of the tension vectors and the weight vector is zero. So the z-component (or the component along the weight direction) of the tension in \( AD \) should balance the weight. Wait, the weight is vertical, so we need the component of \( \mathbf{T}_{AD} \) along the vertical direction (let's say the negative z-direction, so the weight is \( (0,0,-W) \), and the tension in \( AD \) has a z-component of \( T_{AD}\times\frac{26}{55} \)? Wait, no, the vector \( \overrightarrow{AD} \) has components \( (-45, - 18,26) \), so the z-component is 26, x-component is - 45, y-component is - 18. Wait, maybe the vertical direction is the x-direction? No, the crate is hanging, so the weight is vertical, so the vertical direction is along the line from \( A \) to the crate, which is the x-direction (since \( A \) is at (45,0,0) and the crate is at (45,0,0) plus some vertical? Wait, no, the diagram shows the crate is attached at \( A \), and the cables go from \( A \) to \( B \), \( C \), \( D \). Wait, maybe the weight \( W \) is along the negative x-direction? No, that doesn't make sense. Wait, maybe I mixed up the axes. Let's assume that the vertical direction is the direction from \( A \) to the crate, which is the direction opposite to \( AO \), so the length of \( AO \) is 45 in, and the length of \( AD \) is 55 in, so the angle \( \theta \) between \( AD \) and the vertical (AO) is such that \( \cos\theta=\frac{45}{55} \), and the vertical component (along AO) of the tension \( T_{AD} \) is \( T_{AD}\cos\theta \), which should equal the weight \( W \) (since the crate is in equilibrium, the sum of the vertical components of the cables equals \( W \)). Wait, if there are three cables, but maybe in this problem, only the vertical component of \( AD \) is considered, or maybe the other cables have no vertical component? No, that can't be. Wait, maybe the diagram is such that the weight is balanced by the vertical component of \( AD \), and the other cables are horizontal. Wait, the problem says "Determine the weight \( W \) of the crate, knowing that the tension in cable \( AD \) is 924 lb." So maybe the vertical component of \( AD \) is equal to \( W \). So we calculated the vertical component as \( 924\times\frac{45}{55}=756 \) lb. Let's check the calculation: \( 924\div55 = 16.8 \), then \( 16.8\times45 = 756 \). Yes, that's correct.

Answer:

\( \boxed{756} \)