The figure below shows a clawhammer as it is being used to pull a nail out of a horizontal board. A force of magnitude 130 N is exerted horizontally as shown. (NOTE: Assume that the force the hammer exerts on the nail is parallel to the nail and perpendicular to the position vector from the point of contact. The x-component of this force's lever arm is 5 cm.)
(a) Find the force exerted by the hammer claws on the nail.
(b) Find the force exerted by the surface at the point of contact with the hammer head.
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Isn't this just a leaver mechanical advantage issue?
Yes. Still need help figuring out the forces exerted though.
The instructor wants us to use the sum of the torques to figure out the problem.
The web address is posted below my question if you would like to see the picture. The (dot)=.
measure the distance from the hand to the fulcrum, and the distance from the fulcrum to the nail.
Forcehand*distancehandToF=forcenail*distancenailtoF.
Why is this so? The same amount of work is done on each end. Work=forceapplied*distancemoved thru.
However, the distance the force moves through is directly proportional to the distance to the handle fulcurm (thnk on that).
work=torque*angle displacement
now proved by similar triangles the angle displacements are the same on the claw, and the handle. Leaving you with my force*distancetofulcrum argument above.
Thank you very much. That makes sense. I will apply that reasoning.
Did you examine the picture? There are angles involved. The nail is at a 30 degree angle in the wood.
No, didn't see the picture. You take care of that in the torque, sine of the angle between force and the object direction.
To solve this problem, we need to consider the principles of mechanics and use the concept of rotational equilibrium. Let's analyze the forces and torques acting on the hammer.
(a) Finding the force exerted by the hammer claws on the nail:
From the given information, we know that a force of magnitude 130 N is exerted horizontally. Since the force is parallel to the nail, we can assume it acts at a distance of 5 cm from the point of contact.
To find the force exerted by the hammer claws on the nail, we need to balance the torques. The equation for rotational equilibrium is:
Στ = 0
The torque exerted by the force can be calculated using the following formula:
τ = F * r * sin(Θ)
Where:
τ is the torque,
F is the force,
r is the distance from the axis of rotation (point of contact) to the line of action of the force,
Θ is the angle between the force vector and the position vector.
In this case, since the force is horizontal, the angle between the force and the position vector is 90 degrees.
Plugging in the values:
τ = (130 N) * (0.05 m) * sin(90°)
Since sin(90°) equals 1, we can simplify further:
τ = (130 N) * (0.05 m)
τ = 6.5 Nm
The torque exerted by the force is 6.5 Nm. To maintain equilibrium, the torque exerted by the hammer claws on the nail must be equal but in the opposite direction.
Let's assume the force exerted by the hammer claws on the nail is F'. Now we can set up the equation:
(-F') * (0.10 m) = (6.5 Nm)
0.10 m is used instead of 0.05 m because the force exerted by the hammer claws is acting through a longer lever arm.
Solving for F':
F' = (-6.5 Nm) / (0.10 m)
F' = -65 N
Therefore, the force exerted by the hammer claws on the nail is 65 N.
(b) Finding the force exerted by the surface at the point of contact with the hammer head:
To find the force exerted by the surface, we need to realize that the hammer is in rotational equilibrium, which means the sum of all forces acting on it must be zero.
Since the only two forces acting on the hammer are the force exerted by the hammer claws on the nail and the force exerted by the surface at the point of contact, they must cancel each other out.
Considering that the force exerted by the hammer claws on the nail is -65 N (as calculated in part (a)), the force exerted by the surface will be +65 N to produce equilibrium.
Therefore, the force exerted by the surface at the point of contact with the hammer head is 65 N.