Find Weight of Crate

[mathxyz]

Two people are needed to carry a crate. One person exerts a force of 156 pounds and the other person exerts a force of 117 pounds. If the angle between the two forces measures 56.8 degrees, find the weight of the crate to the nearest tenth of a pound.

This question in found in the chapter Parallelogram of Forces and the Law of Cosines. I tried using the law of cosines and then the law of sines trying to guess my way through the question. Help....

 

[Gene]

Sounds like an excellent plan. Why don't you show what you have done????
I hope you did somthing like From O (P(0,0))
Draw a vector OA up and left, length 156.
Draw a vector OB up and right, length 117.
Draw a vector A' from B up and left, length 156 to a point C directly above O.
That is your force diagram. The answer to the question is OC.
Use law of cosines to find AB.
Use law of sines to find <OAB (which is equal to <ABC) and <ABO.
Back to law of cosines to find OC.

 

[soroban]

Hello, mathxyz!

On first reading, it didn't have enough information
. . . then I thought about it . . .

Two people are needed to carry a crate.
One person exerts a force of 156 pounds and the other person exerts a force of 117 pounds.
If the angle between the two forces measures 56.8 degrees,
find the weight of the crate to the nearest tenth of a pound.

          C
          *
         /|\                         The crate is at A.
       /  | \   a
     /    |  \ 156        One person lifts with a force of 156 pounds
 D *      |   \                     in the direction AD.
     \    |b   \
      \   |      * B       The other person lifts with force 117 pounds 
   156 \  |    /                    in the direction AB.
        \ |  / 117
         \|/    c                Angle DAB = 56.8 degrees.
          *
          A

Since we have a parallelogram: .angle B .= .180° - 56.8° .= .123.2°.

Here's the part I had to Think about . . .
If those two people are lifting correctly, the resultant force should be <u>vertical</u>.
. . That is, the resultant (lifting force) is: . b = AC.
. . And the weight of the crate is the opposite vector: -b (pointing down).

Since we're concerned with the <u>magnitude</u> of the vector,
. . we will use the Law of Cosines on triangle ABC.

We have: . a = 156, .c = 117, .<u>/</u> B = 123.2°

. . b² . = . a² + c² - 2ac cos B

. . . . . = . 156² + 117² - 2(156)117)(cos 123.2°) . = . 4441.24791

Hence: . b .= .210.8109293

Therefore, the crate weighs 210.8 pounds.

 

[mathxyz]

hey

Thank you all. Thank you Soroban.