# Thread: First time with vectors

1. ## First time with vectors

Hello everyone,

My name is Alex! I am taking the class algebra 2/ trigonometry honors! I have just finished the algebra 2 section of the class and am moving onto the trigonometry section of the class! Will any of you fine gentlemen be so kind as to help a child like myself with a few simple math problems I cannot understand? Thank you in advance!

12. A force of 176 lbs. Makes an angle of 78°50’ with a second force. The resultant of the two forces makes an angle of 41°10’ with the first force. Find the magnitude of the second force and the resultant.
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13. A force of 28.7 lbs. makes an angle of 42°10’ with a second force. The resultant of the two forces makes an angle of 32°40’ with the first force. Find the magnitude of the resultant.
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14. A crate is supported by two ropes. One rope makes an angle of 46°20’with horizontal and has a tension of
89.6 lbs. on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal
rope.
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15. Two people are carrying a box, one on each side of the box. One person exerts a force of 150 lbs. at an
angle of 62.4° with horizontal. The other person exerts a force 114 lbs. at an angle of 54.9°with horizontal.
Find the weight of the box.

I have the answers to the questions above but I do not understand how to get them! Please show work if you can! Thanks again!
 12. F2:189.59 lbs, r: 282.57 lbs. 13. 116.73 lbs 14. W: 64.8 lbs T: 61.9 lbs 15. 226.2 lbs

2. ## Re: First time with vectors

12. A force of 176 lbs. Makes an angle of 78°50’ with a second force. The resultant of the two forces makes an angle of 41°10’ with the first force. Find the magnitude of the second force and the resultant.
first off, it really helps to make a sketch when solving problems using vectors.

use a coordinate system to your advantage ... let the first force, $F_1 = 176 \, lbs$ , point along the (+) x-axis.

let the second force have magnitude $F_2$ ... its direction will be 78°50’ relative to the (+) x-axis

let the resultant have magnitude $R$ ... its direction will be 41°10’ relative to the (+) x-axis

in the x direction ...

$176 + F_2 \cos(78^\circ 50') = R\cos(41^\circ 10')$

in the y-direction ...

$F_2 \sin(78^\circ 50') = R\sin(41^\circ 10')$

using the second equation, solve for $F_2$ ...

$F_2 = \frac{R\sin(41^\circ 10')}{\sin(78^\circ 50')}$

substitute this expression for $F_2$ in the first equation ...

$176 + \frac{R\sin(41^\circ 10')\cos(78^\circ 50')}{\sin(78^\circ 50')}= R\cos(41^\circ 10')$

$176 = R\cos(41^\circ 10') - \frac{R\sin(41^\circ 10')\cos(78^\circ 50')}{\sin(78^\circ 50')}$

$176 = R\left[\cos(41^\circ 10') - \frac{\sin(41^\circ 10')\cos(78^\circ 50')}{\sin(78^\circ 50')}\right]$

$\frac{176}{\cos(41^\circ 10') - \frac{\sin(41^\circ 10')\cos(78^\circ 50')}{\sin(78^\circ 50')}} = R$

I get $|R| = 286.1 \, lbs$ and $F_2 = 191.9 \, lbs$

I expect round-off error on the part of whoever came up with the provided solutions ... I did not round off in any step of my calculations.

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# a weightless ladder , 20 ft long , rests against a frictionless wall at an angle of 60 degree with tge horizontal . a 150 pound man is 4 ft from the top of the ladder . a horizontal force is needed to prevent it from slipping . choose the correct magnitud

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