Thread: vector analysis problem on dimensions

Suppose a sphere of radius r is dropped into a viscous liquid of coefficient of viscosity q, and its velocity at an instant is v. The frictional force, F, acting against the sphere is given by
F=k(r^x)(q^y)(v^z)
where k is a constant.

Find,using the method of dimensions, the values of x,y,z.
Note the dimensions of the coefficient of viscosity are [M][L]^-1[T]^-1.

My solutions is as follows;

F=ma=[M][L][T]^-2
v=[L][T]^-1

Therefore
[M][L][T]^-2=k[L]^x.q[M]^y.[L]^-y.[T]^-y.[L]^z.[T]^-z

so
[L] 1=X-Y+Z
[M] 1=Y
[T]-2=-Y-Z

and get values x,y,z all equal to 1.

Is this correct? I will appreciate any responses.
Thanks.

Originally Posted by jaspal

Suppose a sphere of radius r is dropped into a viscous liquid of coefficient of viscosity q, and its velocity at an instant is v. The frictional force, F, acting against the sphere is given by
F=k(r^x)(q^y)(v^z)
where k is a constant.

Find,using the method of dimensions, the values of x,y,z.
Note the dimensions of the coefficient of viscosity are [M][L]^-1[T]^-1.

My solutions is as follows;

F=ma=[M][L][T]^-2
v=[L][T]^-1

Therefore
[M][L][T]^-2=k[L]^x.q[M]^y.[L]^-y.[T]^-y.[L]^z.[T]^-z

Extra q left in the RHS of this last equation.

so
[L] 1=X-Y+Z
[M] 1=Y
[T]-2=-Y-Z

and get values x,y,z all equal to 1.

Is this correct? I will appreciate any responses.
Thanks.

Yes this is correct.

RonL