Hammer/Nail problem involving torque

[adamwhite]

Homework Statement

Figure P8.58 shows a claw hammer as it is being used to pull a nail out of a horizontal board. If a force of magnitude 150N is exerted horizontally as shown, find a) the force exerted by the hammer claws on the nail and b) the force exerted by the surface at the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail.

Figure P8.58
http://img137.imageshack.us/img137/2198/hammerbp2.png

Homework Equations

T = rFSin$$\vartheta$$

The Attempt at a Solution

Using trig I find that the force the prongs exert on the nail is 300N

 

[Oerg]

you should check your answer.the torque caused by the horizontal force is 45. This means that the vertical force exerted is 900.

 

[adamwhite]

How did you get your answers?

 

[Oerg]

150x0.3/0.05?

 

[adamwhite]

Why do you do it that way instead of the way I did it? Can you explain that please?

 

[Oerg]

Well, how did you do it? The vertical force exerted on the nail is 900N. The nail needs to exert a component that would equal 900N. The toher component turns up as the contact force 9well part of, not forgeting the weight of the hammer).

 

[adamwhite]

This is how I did it:

My diagram looks like this
http://img219.imageshack.us/img219/7730/nailod0.png

Since the 150N is the horizontal component of the force exerted on the nail I used trig to find the force.

sin30 = 150/F
F = 150/sin30 = 300

 

[Oerg]

you should do the question via moments.

Imagine you had a longer handle, wouldn' it be easier to retrieve the nail?

 

[adamwhite]

Yeah, you're right it would. But I still don't see how you got your numbers :(

 

[Oerg]

the torque caused by the hammer is 150 x 0.3.

if the nail is not moving, then it is countering this torque with its own. So to counter this, it would need to exert a force which has a component that would counter this torque.

EDIT: i edited my last few posts for clarity

 

[Oerg]

by the way my previous number of 788 was wrong it should be higher than 900

 

[ganaa]

I tried it by moments as oerg suggested:

Taking moments about the point of contact of the hammer and the floor (this eliminates the reaction of the floor).
In equilibrium:

moment of effort on hammer =150*0.3 = moment of hammer on nail force= F*(0.05/cos(30 deg))

solving for F gives 780 N.