Math / Geometry

Question36. (II) Two large snowcats are towing a housing unit north, as shown in Fig. 4-42. The sum of the forces $\overrightarrow{\mathbf{F}}_{\mathrm{A}}$ and $\overrightarrow{\mathbf{F}}_{\mathrm{B}}$ exerted on the unit by the horizontal cables is north, parallel to the line L, and $F_{\mathrm{A}}=4200 \mathrm{~N}$ . Determine $F_{B}$ and the magnitude of $\overrightarrow{\mathbf{F}}_{\mathrm{A}}+\overrightarrow{\mathbf{F}}_{\mathrm{B}}$ .
FIGURE 4-42 Problem 36.

Studdy Solution

STEP 1

1. The forces $\overrightarrow{\mathbf{F}}_{\mathrm{A}}$ and $\overrightarrow{\mathbf{F}}_{\mathrm{B}}$ are applied at angles to the north line $L$ .
2. The sum of the forces is directed exactly north, parallel to line $L$ .
3. The angle for $\overrightarrow{\mathbf{F}}_{\mathrm{A}}$ is $48^\circ$ to the north line.
4. The angle for $\overrightarrow{\mathbf{F}}_{\mathrm{B}}$ is $32^\circ$ to the north line.
5. $F_{\mathrm{A}} = 4200 \, \text{N}$ .

STEP 2

1. Resolve the forces into their northward (vertical) and eastward (horizontal) components.
2. Set up equations based on the condition that the sum of the forces is directed north.
3. Solve for $F_{\mathrm{B}}$ .
4. Calculate the magnitude of $\overrightarrow{\mathbf{F}}_{\mathrm{A}} + \overrightarrow{\mathbf{F}}_{\mathrm{B}}$ .

STEP 3

Resolve the forces into their components:
For $\overrightarrow{\mathbf{F}}_{\mathrm{A}}$ : - Northward component: $F_{\mathrm{A}} \cos(48^\circ)$ - Eastward component: $F_{\mathrm{A}} \sin(48^\circ)$
For $\overrightarrow{\mathbf{F}}_{\mathrm{B}}$ : - Northward component: $F_{\mathrm{B}} \cos(32^\circ)$ - Eastward component: $F_{\mathrm{B}} \sin(32^\circ)$

STEP 4

Set up equations based on the condition that the sum of the forces is directed north:
Since the sum of the horizontal components must be zero (as the resultant force is purely northward):
$F_{\mathrm{A}} \sin(48^\circ) = F_{\mathrm{B}} \sin(32^\circ)$

STEP 5

Solve for $F_{\mathrm{B}}$ :
$F_{\mathrm{B}} = \frac{F_{\mathrm{A}} \sin(48^\circ)}{\sin(32^\circ)}$
Substitute $F_{\mathrm{A}} = 4200 \, \text{N}$ :
$F_{\mathrm{B}} = \frac{4200 \times \sin(48^\circ)}{\sin(32^\circ)}$
Calculate $F_{\mathrm{B}}$ :
$F_{\mathrm{B}} \approx \frac{4200 \times 0.7431}{0.5299}$ $F_{\mathrm{B}} \approx 5896.3 \, \text{N}$

STEP 6

Calculate the magnitude of $\overrightarrow{\mathbf{F}}_{\mathrm{A}} + \overrightarrow{\mathbf{F}}_{\mathrm{B}}$ :
The northward components add up to give the resultant force:
$F_{\text{resultant}} = F_{\mathrm{A}} \cos(48^\circ) + F_{\mathrm{B}} \cos(32^\circ)$
Substitute the values:
$F_{\text{resultant}} = 4200 \times \cos(48^\circ) + 5896.3 \times \cos(32^\circ)$
Calculate $F_{\text{resultant}}$ :
$F_{\text{resultant}} \approx 4200 \times 0.6691 + 5896.3 \times 0.8480$ $F_{\text{resultant}} \approx 2810.2 + 4999.7$ $F_{\text{resultant}} \approx 7810 \, \text{N}$
The magnitude of $\overrightarrow{\mathbf{F}}_{\mathrm{A}} + \overrightarrow{\mathbf{F}}_{\mathrm{B}}$ is:
$\boxed{7810 \, \text{N}}$

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