Let's solve **step by step**: --- ## **Step 1: Understand the Problem** - A 19-kg flowerpot is suspended by three cords at point \( A \). - The cords are \( AB \), \( AC \), and \( AD \). - We need to find the tensions (forces) in the cords for equilibrium. --- ## **Step 2: Draw and Assign Coordinates** From the figure (and typical conventions), let's assign coordinates: - \( A \): Origin \((, , )\) - \( B \): Along \( y \), \( 48^\circ \) above the \( xy \)-plane. - \( C \): In the \( xz \)-plane, \( 30^\circ \) from \( z \), \( 30^\circ \) from \( x \). - \( D \): In the \( xz \)-plane, \( 60^\circ \) from \( z \), \( 30^\circ \) from \( x \). --- ## **Step 3: Find Direction Vectors** ### **Cord \( AB \):** - Makes a \( 48^\circ \) angle with the \( y \)-axis (upwards). - Vector: \( (, \cos48^\circ, \sin48^\circ) \) ### **Cord \( AC \):** - In the \( xz \)-plane, \( 30^\circ \) from \( z \)-axis. - Vector: \( (\sin30^\circ, , \cos30^\circ) \) ### **Cord \( AD \):** - In the \( xz \)-plane, \( 60^\circ \) from \( z \)-axis (or \( 30^\circ \) from \( x \)-axis, but negative \( x \) direction). - Vector: \( (-\sin30^\circ, , \cos30^\circ) \) --- ## **Step 4: Write Force Equilibrium Equations** Let the tensions be \( F_{AB} \), \( F_{AC} \), \( F_{AD} \). The sum of all forces must be zero: \[ \vec{F}_{AB} + \vec{F}_{AC} + \vec{F}_{AD} + \vec{W} = \] Where \( \vec{W} = (,,-mg) \) and \( m = 19\,\text{kg} \), \( g = 9.81\,\text{m/s}^2 \). \[ mg = 19 \times 9.81 = 186.39\,\text{N} \] --- ## **Step 5: Resolve Each Force into Components** ### **Unit Vectors for Each Cord** #### \( \vec{u}_{AB} \): \[ = (, \cos48^\circ, \sin48^\circ) = (, .6691, .7431) \] #### \( \vec{u}_{AC} \): \[ = (\sin30^\circ, , \cos30^\circ) = (.5, , .866) \] #### \( \vec{u}_{AD} \): \[ = (-\sin30^\circ, , \cos30^\circ) = (-.5, , .866) \] --- ### **Write Components for Each Force** Let \( F_{AB}, F_{AC}, F_{AD} \) be the tensions. - \( \vec{F}_{AB} = F_{AB} (, .6691, .7431) \) - \( \vec{F}_{AC} = F_{AC} (.5, , .866) \) - \( \vec{F}_{AD} = F_{AD} (-.5, , .866) \) --- ### **Sum Forces in Each Direction** #### **x-direction:** \[ .5 F_{AC} - .5 F_{AD} = \implies F_{AC} = F_{AD} \] #### **y-direction:** \[ .6691 F_{AB} = \implies F_{AB} = \] But this can't be correct (since the system needs 3 cords). Let's check the geometry: The problem is, all the y-components are from cord \( AB \), and the other two are in the xz-plane. So this is actually correct: only \( AB \) provides the y-component to balance the weight. #### **z-direction:** \[ .7431 F_{AB} + .866 F_{AC} + .866 F_{AD} - 186.39 = \] But since \( F_{AC} = F_{AD} \): \[ .7431 F_{AB} + 2 \times .866 F_{AC} = 186.39 \] --- ## **Step 6: Solve the System** ### **x-direction:** \[ F_{AC} = F_{AD} \] ### **y-direction:** \[ .6691 F_{AB} = \implies F_{AB} = \] But, the cords must be in tension. Let's re-express the geometry for \( AB \): If \( AB \) is in the \( y \)-\( z \) plane, then its vector is \( (, \cos 48^\circ, \sin 48^\circ) \). The flowerpot is supported, so the y-component doesn't need to be zero. Let's proceed with: #### y-direction: \[ .6691 F_{AB} = \] So, y-direction must be zero net force. But gravity acts only in the z-direction. Hence, the only nonzero y-component comes from \( AB \). But the pot is not being pulled in the y-direction, so the net y-force must be zero. Therefore, \( F_{AB} \) is not zero, but it doesn't balance any force in y. Let's focus on z and x. #### x-direction: \[ .5 F_{AC} - .5 F_{AD} = \implies F_{AC} = F_{AD} \] #### z-direction: \[ .7431 F_{AB} + .866 F_{AC} + .866 F_{AD} = 186.39 \] \[ .7431 F_{AB} + 1.732 F_{AC} = 186.39 \] #### y-direction: \[ .6691 F_{AB} = \implies F_{AB} = \] But \( F_{AB} = \) would mean we only have 2 cords, which isn't realistic. So, let's check the geometry again. #### Let's check the direction vectors for the cords: - **AB:** In \( yz \)-plane at \( 48^\circ \) above \( y \)-axis. - \( AB = (, 1, ) \) to \( (, \cos48^\circ, \sin48^\circ) \) - **AC:** Makes \( 30^\circ \) with \( z \)-axis in \( xz \)-plane. \( (\sin30^\circ, , \cos30^\circ) \) - **AD:** Makes \( 60^\circ \) with \( z \)-axis in \( xz \)-plane. \( (\sin60^\circ, , \cos60^\circ) \) So, let's correct: - \( AC = (a, , b) \) where \( a = \sin 30^\circ = .5, b = \cos 30^\circ = .866 \) - \( AD = (-\sin 30^\circ, , \cos 30^\circ) = (-.5, , .866) \) So our previous vectors were correct. Now, **y-direction**: Only \( AB \) provides a y-component. So for equilibrium, **no external force in y**, so \[ .6691 F_{AB} = \implies F_{AB} = \] So the tensions in \( AC \) and \( AD \) must support the entire weight, and \( AB \) is slack. --- ## **Final Calculations** ### **x-direction:** \[ .5 F_{AC} - .5 F_{AD} = \implies F_{AC} = F_{AD} \] ### **z-direction:** \[ .866 F_{AC} + .866 F_{AD} = 186.39 \] \[ 1.732 F_{AC} = 186.39 \implies F_{AC} = \frac{186.39}{1.732} = 107.6\,\text{N} \] \[ F_{AD} = F_{AC} = 107.6\,\text{N} \] ### **Summary of Answers** #### **Part A: \( F_{AB} \)** \[ \boxed{\,\text{N}} \] #### **Part B: \( F_{AC} \)** \[ \boxed{107.6\,\text{N}} \] #### **Part C: \( F_{AD} \)** \[ \boxed{107.6\,\text{N}} \] --- ## **Step-by-Step Summary Table** | Cord | Force (N) | |--------|-----------| | AB | | | AC | 107.6 | | AD | 107.6 | --- ### **Explanation** - The geometry and equilibrium require only two cords to support the entire weight, and the third is slack. - If the geometry is interpreted differently or if the image details differ, let me know, and I can recalculate with more precise vectors.