Angle between a force and radius?

[TexasCow]

Homework Statement

A force, F=(6i-2j)[N], moves a particle through a distance deltaR=(3i+1j)[m]

A. Find the work done on the particle

B. Find the angle between F and deltaR

Homework Equations

W=F*D

The Attempt at a Solution

A. F=Root(6^2+2^2)=6.32N

R=Root(3^2+1^2)=3.16m

W=FD=(6.32*3.16)=20.00N/M

I can't figure out B though. By the way, is my A correct? Thanks!

 

[rl.bhat]

W=FD=(6.32*3.16)=20.00N/M It is true if they are in the same direction.
W= F.deltaR = (6i-2j).(3i+1j) = 18 - 2 = 16J
And angle is given by cos(theta) = F.d/FD

 

[TexasCow]

I'm kind of confused. Did you foil those out? I was taught to use pythag. theorum for component vectors. I'm also kind of confused on F.D/FD. What does this mean?

 

[kudoushinichi88]

I think B can be found if you find the dot product of the two vectors...

 

[rl.bhat]

In vector you can perform scalar product and vector product. If Fand D are the force and dispalcement vectors then the work done = F.D and angle between the vectors =F.D/FD Bold letters indicatre the vectors.

 

[kudoushinichi88]

Oh, so the dot product is called the scalar product around here...

 

[TexasCow]

Oh I think I see what you're saying. So the scalar, or dot product divided by FD is equal to the angle?

 

[rl.bhat]

Yes.

 

[TexasCow]

I think I just confused myself. You said W=F.D, or F multiplied by D. Then, you say that the angle is F.D/FD. If F.D is F multiplied by D, then wouldn't F.D/FD be 1? Since top and bottom would be the same? Or is the top the dot product and and bottom FxD?

Also, the dot product is:

lAl*lBlcosx correct?

 

[rl.bhat]

F.D = 6i-2j).(3i+1j) = 18 - 2 = 16J It is dot product.
FD = FD=(6.32*3.16)=20.00N/M
And cosx = 18/20

 

[carloranosa]

radius

the diameter of the large and the small piston of a hydraulic lift

If the ratio of the diameter of the small piston to that of the large piston is 1:2, how does the forco on the small piston compare with that bon the large piston?

a hydraulic is to be constructed so that a force of 10 Newton can lift a load of 6250 Newton. if the small piston has a radious of 4 cm, what must be the radious of the large piston?

 

[carloranosa]

radius

can please somebody help me solve these fallowing problems?

the area of the small and large pistons of a hydraulic lift are 10cm and 500cm, respectively. What load can be lifted on the large piston if a force of 40 Newton is applied in the small piston?

the diameter of the large and the small piston of a hydraulic lift are 6cm and 75cm, respectively. what force must be applied on the piston if a load of 4000 Newton must be overcome by the large piston?

If the ratio of the diameter of the small piston to that of the large piston is 1:2, how does the forco on the small piston compare with that bon the large piston?

a hydraulic is to be constructed so that a force of 10 Newton can lift a load of 6250 Newton. if the small piston has a radious of 4 cm, what must be the radious of the large piston?