DWelsh Posted February 14, 2017 Share Posted February 14, 2017 I'm taking an engineering intro class, and this problem has me stuck: "Vector R in Fig. 4.9 is the difference between vectors T and S. If S is inclined at 28 degrees from the vertical and the angle between S and T is 32 degrees, calculate the magnitude and direction of vector R. The magnitude of S and T are 19 cm and 36 cm, respectively. " Fig. 4.9 Is simply three lines pointing diagonally downward from the top right, with R being horizotal, T below R, and S below T (sorry, I cant find a picture to upload). I only have pre-calc under my belt (no physics) but I feel like this shouldn't be to hard to understand. I'm just finding it impossible to figure out HOW to solve this. I have access to the answer:(R = 18.8 cms, direction is horizontal) but I really need to know how to do it. Any help is greatly appreciated. Thanks Link to comment Share on other sites More sharing options...
zztop Posted February 14, 2017 Share Posted February 14, 2017 (edited) Tr I'm taking an engineering intro class, and this problem has me stuck: "Vector R in Fig. 4.9 is the difference between vectors T and S. If S is inclined at 28 degrees from the vertical and the angle between S and T is 32 degrees, calculate the magnitude and direction of vector R. The magnitude of S and T are 19 cm and 36 cm, respectively. " Fig. 4.9 Is simply three lines pointing diagonally downward from the top right, with R being horizotal, T below R, and S below T (sorry, I cant find a picture to upload). I only have pre-calc under my belt (no physics) but I feel like this shouldn't be to hard to understand. I'm just finding it impossible to figure out HOW to solve this. I have access to the answer:(R = 18.8 cms, direction is horizontal) but I really need to know how to do it. Any help is greatly appreciated. Thanks Triangle cosine rule teaches us that: [latex]R^2=S^2+T^2-2S*T*cos x[/latex] where x is the angle between S and T. I'll let you figure why R is horizontal Edited February 14, 2017 by zztop 1 Link to comment Share on other sites More sharing options...
DWelsh Posted February 14, 2017 Author Share Posted February 14, 2017 Thank you for your help. I can see that this is just simple trigonometry now. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now