Question

Can someone help me check my answer to these three questions and help me with one question?

Answer 1

@Mertsj

Answer 2

Your answers look correct to me

Answer 3

What about the final question? I don't get it. @rauber101

Answer 4

To solve this, you need to break each force into a horizontal and vertical component using trig.

Answer 5

How?

Answer 6

Look at each as a triangle:

For example the 200 lb force at 60 degrees can be broken down

F=<x,y>
horizontal component (x) = 200cos(60)
vertical component (y) = 200sin(60)

Answer 7

Then do the same for the 100lb force, then add the two together?

Answer 8

Exactly

Answer 9

The only problem I see is that I'm not sure if the forces are supposed to be acting towards the object or away from it: This would change all the numbers so they are the opposite direction

Answer 10

I'm just gonna assume that it's acting towards it. Is this correct?:

Direction of the first force=<200*cos(60), 200*sin(60)>=<200*0.5, 200*0.87>=<100, 174>
Direction of the second force=<100*cos(170), 100*sin(170)>=<100*-0.98, 100*0.17>=<-98, 17>
Resultant force's direction=<100-98,174+17>=<2,191>
Resultant force's magnitude=sqrt(2^2+191^2)=sqrt(4+36481)=sqrt(36485)=191.01 lbs

Answer 11

@rauber101

Answer 12

That looks correct to me

Answer 13

Thanks!