One More Physics/Trig Question For You Guys

I would personally treat this as a vector math problem, arbitrarily letting truck 1 pull in the X direction. Give this a whirl and if you're still stuck, I can help you a bit further.

 

You could also draw a trapezoid whose two sides are the two forces, separated by the angle, and then find the distance from corner to corner of the trapezoid, but here is just my personal opinion: Vectors were invented to make trig easier.

 

I'd draw a free body diagram, then turn that into a vector addition triangle to be solved using the cosine rule. HINT IN WHITE FONT, HIGHLIGHT TO VIEW- the triangle will have two sides 10980 and 12450 units long with an angle between them of 151.5 degrees. You need to find the length of the 3rd side.

 

Yeah vector math, find forces in the x direction. Forces in the y direction will be the tow trucks pulling on each other

 

Use my hint!

Forget x and y stuff, it's irrelevant. To add vectors you draw them end to end. Then use the cosine rule. You will get ONE answer that will be the resultant force on the truck. I make it 22712 N.



NOT DRAWN TO SCALE!!

 

yeah,,this is correct. unless it's a trick question (duh,,all forces acting on the truck)

 

These are making my head hurt. Can we have more funny photos please? kthxbai

 

Obviously you need to perform this experiment in the field to get the proper results. Aquisition of two tow-trucks and large truck is neccesary, as well as a ditch.

 

It's fun to look at different ways to solve these problems. In my old age, I use a spreadsheet for practically all math, so my computations get done in a whole lot of little steps. I'm going to create vectors for both of the forces. The first one I will arbitrarily orient along the X axis:

F1x = 10980
F1y = 0

The other one has components determined by the sine and cosine of the angle:

F2x = 12450*cos(28.5) = 10941
F2y = 12450*sin(28.5) = 5940

Now, remember that force vectors simply add:

Fx = F1x + F2x = 10980 + 10941 = 21921
Fy = F1y + F2y = 0 + 5940 = 5940

The magnitude of a vector is found by the Pythagorean theorem:

F = Sqrt(Fx^2 + Fy^2) = 22711 Newtons

Give or take a decimal place.

The law of cosines is certainly a quicker way to solve this, if you can remember the law of cosines. However, the laws of trig get really hairy in three dimensions, and vector math wins the race.

 

If you come across a car in the ditch don't do anything. You'll get sued.

 

Vectors have to be added "head to tail". What you've done is put both tails together, and that creates a nonsensical answer. Bassybill has the fastest way of solving the question. It's simple vector manipulation followed by the cosine rule.

 

Okay, I've added some more to my diagram.





YOUR triangle is based on the forces acting on the truck in the ditch. You will probably see that it's similar to what I've drawn on the left of my picture. On that side, you can see the forces from the two tow trucks acting on the truck in the ditch (not shown, but obviously at the bottom of the pic). We need to add these two forces. To do so, we "slide" the one on the right along the other force, up and to the left, so the forces are now "head-to-tail". Hopefully you'll be able to see that the angle between the right hand force and the dotted line extension of the other force is still 28.5 degrees. This means the angle between the two forces in the vector addition triangle is 180-28.5=151.5 degrees. Now it's just a question of using the cosine rule to find the resultant (3rd side of the triangle) and then sine rule/subtraction to get the other angles if you want them - the question doesn't actually ask for these.

fdeck's solution is effectively doing exactly the same thing, mathematically speaking - just a different approach to the same problem. By the way, what value did you have provided as the solution? I'm curious as to whether this one was actually correct, as for both the other two problems you posted the answers were wrong.

 

Okay, I'll admit it's been almost 40 years since I did a trig problem so I probably made a stupid mistake here - perhaps someone will spot it quickly. I believe the formula you are using is c[SUP]2[/SUP] = a[SUP]2[/SUP] + b[SUP]2[/SUP] - 2abcos(angle) where a = 10980, b= 12450 and the angle is 151.5. However, when I plugged all this into a spreadsheet I got a[SUP]2[/SUP] = 15502500, b[SUP]2[/SUP] = 275562900, and 2ab = 273402000 and cos(151.5) = .7625349 which gave me a solution of 67080560.3 and the square root being 8190.2723 which is way off.

 

Are you using the radians mode on your calculator, while punching in degrees?

EDIT: You most definitely are.

FWIW Second sector angles (between 90 to 180 degrees) are always negative. Recall the cosine curve. Also, Excel and Google Calc work in radians by default.

 

Thanks.