Hi
the force from the hammer is 5,4 * F according to my book .. How did they calculate that ? Should not it be 27 * F
thanks
[Physics] determine force from a hammer to pull out a nail
forceshomework-and-exercisesnewtonian-mechanics
Related Solutions
An object of mass $m$ on an incline with friction experiences the following three forces:
- Gravity
- Normal force
- Friction
The normal force points away from the incline's surface and is perpendicular to it. The force of friction is parallel to the incline and points in the direction opposing the motion of the object (in this case that means it points up the incline). The gravitational force points down (in the direction of the negative $y$-axis).
Since the object is not accelerating in the direction perpendicular to the incline (otherwise it would be falling through the surface or losing contact with the surface), one concludes that the component of the gravitational force perpendicular to the incline cancels the normal force, and the net force in that direction is zero.
What remains is the component of the gravitational force pointing down the incline, and the friction force pointing up the incline. A picture and some triangle-drawing/trig should convince you that the magnitude of the component of the gravitational force pointing down the incline is
$m g \sin (30^\circ) = (1\,\mathrm{kg})(9.8\,\mathrm{m}/\mathrm{s}^2)(1/2) = 4.9\,\mathrm N$.
The friction force pointing up the incline has magnitude $1.5\,\mathrm N$. Since the component of the gravitational force along the incline points down the incline, and the friction force points up the incline, the magnitude of the net force is just
$4.9\,\mathrm N - 1.5\,\mathrm N = \boxed{3.4\,\mathrm N}$.
Let me know if anything here was confusing, and I can make more comments.
Cheers!
Of course the force changes during the impact - so to get close to an answer, you need both the time of the impact and the magnitude of the momentum transfer.
As user77567 pointed out, a fairly simple way to measure momentum transfer is with a ballistic pendulum. This would be a heavy steel ball (much heavier than the hammer) hung from a long wire. When you strike the ball, the hammer will bounce back (since it is much lighter) and the ball will swing through an arc $\alpha$ before returning to the equilibrium position.
If ball has mass $M$ and hammer mass $m<<M$, then conservation of momentum tells us that
$$M\cdot v_{ball}= m\cdot \Delta v_{hammer}$$
For an elastic collision with $m<<M$, $\Delta v_{hammer} \approx 2 v_{initial}$
If the wire has length $\ell$ and moves through a distance $d$, so $\alpha = \tan^{-1} \frac{d}{\ell}$, conservation of energy tells us that for small deflections, the height $h$ that the ball rises after the impact is
$$h = \ell (1-\cos\alpha)\approx \frac{d^2}{2\ell}$$
Conservation of energy then tells us
$$M\cdot g\cdot h = \frac12 M v^2$$
and it follows that
$$v = d\sqrt{\frac{g}{\ell}}$$
Of course we could have got the same result directly from the equation of motion for a simple harmonic oscillator (pendulum).
The remaining interesting question is the impact time. This can be measured with simple electronic components. If you connect a resistor and a charge capacitor in parallel, with a "switch" formed by the contact between the hammer and the ball, then you can compute the impact time by observing the fraction of discharge of the capacitor due to the "closing of the switch" when the hammer hits the ball. Sufficiently thin and flexible wires should allow this measurement without disturbing the mechanics. Use a digital multimeter with sufficiently high impedance (at least 10 M). If the capacitor leaks slowly after you first charge it (say with a battery), you can observe the voltage dropping and hit the ball with the hammer just as the voltage hits a "round" value - this allows you to minimize the ffect of drift.
To make the measurement of impact time repeatable you could make the hammer part of a second pendulum that hits the ball from different heights: you can then plot the relationship between impact velocity and impact time, and this will allow you to get the time when you hit the ball really hard (when you might not get a good repeatable measurement of the time or velocity).
I hope this is enough to get you going...
Best Answer
Do a torque balance around the point of rotation. The applied force gives a torque of $27*F$. The nail gives a torque of $-5*f_{nail}$. These torques must sum to zero, so solving for the nail force gives $f_{nail}=(27/5)F=5.4*F$