Equilibrium Ball attached to wall problem

[brendan3eb]

Homework Statement

A uniform sphere of mass m and radius r is held in place by a massless rope attached to a frictionless wall a distance L above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.

Homework Equations

net torque = 0
net force = 0
t=force of tension
F=force of wall
mg=gravitational force on ball

The Attempt at a Solution

well, strangely enough I am able to solve for part B. choosing the point where the string is attached to the wall as my origin. I am able to determine the torque caused by tension to be 0 as the tension force runs 180 parallel the string. Using motion arm times force = torque, I can find the remaining torques so that:
-mgr+t*0+L*F=0
which yields
F=mgr/L
which is the correct answer to part B. However, I have no clue how to solve part A. I have already used my torque, so I tried setting the force equations for the x and y-axis to 0.
T*cos(x)-mg=0
F-T*sin(x)=0
but that doesn't get me anything close to the correct answer of
T=(mg/L)*sqrt(L^2+r^2)

help please

 

[Doc Al]

However, I have no clue how to solve part A. I have already used my torque, so I tried setting the force equations for the x and y-axis to 0.
T*cos(x)-mg=0
F-T*sin(x)=0
but that doesn't get me anything close to the correct answer of
T=(mg/L)*sqrt(L^2+r^2)

Either one of those force equations will get you the answer. Hint: Evaluate sin(x)--or cos(x)--in terms of the lengths given.

 

[brendan3eb]

oh, thanks!...so

t*cosx-mg
t*cosx=mg
replace L/a for cosx
t*(L/a)=mg
t=mg/L*a
by pyth. theorem a = sqrt(r^2+L^2)
t=(mg/L)sqrt(r^2+L^2)

THANK YOU SO MUCH! It feels so good to get that problem done. For awhile I thought that there would be something way complex to do that would take a lot of work...and I was intimidated to go any further. Thanks again.