Vectors (1 Viewer)

[=)(=]

Q11 pls

 

[ExtremelyBoredUser]

Q11 pls

Take up/right as positive and refer to diagram. The diagram is both the forces so you can just break it down to horizontal and vertical components and add them together respectively to get the resultant force components.



from reading of the diagram;



do the calculation for net force and direction and you should get
above the horizontal/above east (to 1 d.p)

I think the expected way is finding the components for each force vector, listing them and them adding them separately or just using the matrices form to add them together and then you would arrive at the same step but cant be bothered doing all.

 

[5uckerberg]

Take up/right as positive and refer to diagram. The diagram is both the forces so you can just break it down to horizontal and vertical components and add them together respectively to get the resultant force components.

View attachment 34756

from reading of the diagram;



do the calculation for net force and direction and you should get
above the horizontal/above east (to 1 d.p)

I think the expected way is finding the components for each force vector, listing them and them adding them separately or just using the matrices form to add them together and then you would arrive at the same step but cant be bothered doing all.

From memory to find the direction what should be done is to move Nitro's arrow up to where Brutus's arrow is so that you get something like this

 

[5uckerberg]

Once you have that a cosine rule for the last side and then a sine will be done on the big angle note you have to add 90 degrees into your calculations after doing the inverse sine and then take away 135 degrees because by doing so one can see the resultant force.

Note what I said is actualy a mistake instead the correct way to find the direction is that once you know how much net force is there horizontally and vertically you then do which gives us 5.1626... degrees which are simply 5.2 degrees rounded to 2 significant figures. @ExtremelyBoredUser I guess wanted to do it the formal way but I think my idea is much easier for people to do.

 

[Jojofelyx]

Take up/right as positive and refer to diagram. The diagram is both the forces so you can just break it down to horizontal and vertical components and add them together respectively to get the resultant force components.

View attachment 34756

from reading of the diagram;



do the calculation for net force and direction and you should get
above the horizontal/above east (to 1 d.p)

I think the expected way is finding the components for each force vector, listing them and them adding them separately or just using the matrices form to add them together and then you would arrive at the same step but cant be bothered doing all.

neat

 

[=)(=]

Got it thanks guys

 

[=)(=]

wait for some reason

Once you have that a cosine rule for the last side and then a sine will be done on the big angle note you have to add 90 degrees into your calculations after doing the inverse sine and then take away 135 degrees because by doing so one can see the resultant force.

Note what I said is actualy a mistake instead the correct way to find the direction is that once you know how much net force is there horizontally and vertically you then do which gives us 5.1626... degrees which are simply 5.2 degrees rounded to 2 significant figures. @ExtremelyBoredUser I guess wanted to do it the formal way but I think my idea is much easier for people to do.

How did you know that it was going to be 5.2 from the horizontal?

 

[5uckerberg]

wait for some reason

How did you know that it was going to be 5.2 from the horizontal?

Remember the net forces horizontally and vertically line them up so that you have a right angled triangle and then use the inverse tan as shown from what I said for the opposite which is the vertical length and then adjacent is the horizontal length.

Next, we just do [/TEX]

 

[=)(=]

ohh i see thanks