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April 21, 2014

April 21, 2014

Posted by **HALP** on Friday, November 30, 2012 at 5:07pm.

(a) Find the magnitude of the resulting acceleration of the center of mass of the stick.

(b) While the 12.7-N force is being applied, find the horizontal force on the stick by the hinge.

(c) There is some point at which the hammer can strike the stick such that the horizontal force on the stick by the hinge will be zero, even though the stick is being struck by the hammer at that instant. Calculate the distance from the hinge to the location of this special point.

At__________m below the hinge.

- physics -
**Damon**, Friday, November 30, 2012 at 5:39pmM=Torque about hinge = 12.7 * .75 = 9.525

I about hinge = (1/3) m l^2 = (.68/3)2.25

= .51

angular acceleration alpha = M/I = 9.525/.51 = 18.7 radians/s^2

resulting linear acceleration of center East = .75 alpha = 14 m/s^2 (part a)

we better get the same acceleration East from the sum of forces on the stick

12.7east + force from hinge F = .68 (14)

12.7 + F = 9.52

F = -3.18 or 3.18 N west

Now the sweet point:

Torque

M = 12.7 x

alpha = 12.7 x/.51 = 24.9 x

a of center = .75 alpha = 18.7 x meters/s^2

F = m a

18.7 x = F/m but F is just 12.7 because the hinge force is zero so

18.7 x = 12.7/.68

x = 1 meter, the end of the stick

Try this with a mass distribution like a baseball bat or tennis racquet and the result will make more sense :)

- physics -
**HALP**, Friday, November 30, 2012 at 5:46pmthanks

- physics -
**Damon**, Friday, November 30, 2012 at 7:20pmOh, one meter is not the end of the stick. It is 2/3 of the way out.

- physics -
**Damon**, Friday, November 30, 2012 at 7:23pmWhich is about how far out on a bat you want to hit the ball without stinging your hand.

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