What is the force exerted on a stick by a bowling ball on a rough surface?

[proace360]

Homework Statement

A stick with a mass of 0.205 kg and a length of 0.432 m rests in contact with a bowling ball and a rough floor, as shown in the figure below. The bowling ball has a diameter of 21.9 cm, and the angle the stick makes with the horizontal is 30°. You may assume there is no friction between the stick and the bowling ball, though friction with the floor must be taken into account.

(a) Find the magnitude of the force exerted on the stick by the bowling ball.
(b) Find the horizontal component of the force exerted on the stick by the floor.
(c) Repeat part (b) for the upward component of the force.

Homework Equations

Torque=Nm

The Attempt at a Solution

You can tell I am having some trouble with Physics today . So, I'm trying to do a), but I can't think of a way to find how far away on the stick the bowling ball makes contact :/. I've tried EVERYTHING but I can't figure it out.

 

[Rake-MC]

30 degree incline. So perpendicular to the incline will make 60 degrees with the horizontal.
Imagine a horizontal line straight through the middle of the ball, The 60 degree line going through the middle also. You can see this appears to be like the unit circle.

You can see the perpendicular line we just made corresponds with [tex] \frac{2\pi}{3} [/tex]

calculate the height at this point. HINT: radius = 0.1095m, so height is [tex] 0.1095sin(\frac{\pi}{3}) [/tex]

Add this height to radius of ball = height of stick where ball is in contact.

and then just divide that given value by sine of 30 degrees. Voila!

That's probably the long way of doing it, but at least you understand how it works.

 

[proace360]

Sorry, I am still confused... how does that give the height of the ball? The points of contact with the floor and stick do not lie on a line going the through the center of the ball (if that makes sense)

 

[Rake-MC]

Okay, so what I've done is calculate the VERTICAL distance from the centre of the ball to the part that it's touching. (0.1095sin(60)).

That is, if you draw a horizontal line that goes straight through the dead centre of the ball, and then drop a vertical line from where the ball and stick intersect down to the horizontal line, the length of this vertical line will be 0.1095sin(60).

However, the other end of the stick touches the ground. Therefore we need to add the vertical distance from the horizontal imaginary line to the ground, which conviniently is r.
So the distance from where the ball intersects the stick, to the ground = 0.1095sin(60) + 0.1095

Use trigonometry to find the hypotenuse.