To find the magnitude of the force exerted by the hammer claws on the nail, we can use the principle of moments (torque) about the point of contact between the hammer and the board. The torque due to the horizontal force (125 N) is equal to the torque due to the force exerted by the hammer claws.
Let F be the force exerted by the hammer claws on the nail.
Torque due to horizontal force = Torque due to force by hammer claws
125 N * 30 cm * sin(32.2°) = F * 5 cm
Solve for F:
F = (125 N * 30 cm * sin(32.2°)) / 5 cm
F ≈ 201.83 N
The magnitude of the force exerted by the hammer claws on the nail is approximately 201.83 N.
To find the direction of the force exerted by the hammer claws, we need to find the angle between the force and the nail. This angle can be found using the arctangent function. Since the contact point is 30 cm from the force and the head of the nail is 5 cm from the contact point, we have:
tan(angle) = (5 cm) / (30 cm)
angle = arctan(5/30)
angle ≈ 9.46°
The direction of the force exerted by the hammer claws on the nail is approximately 9.46° from the nail.
The figure below shows a claw hammer being used to pull a nail out of a horizontal board where θ = 32.2°. The mass of the hammer is 1.00 kg. A force of 125 N is exerted horizontally as shown, and the nail does not yet move relative to the board. Assume the force the hammer exerts on the nail is parallel to the nail. the contact point is 30 cm from the force and the head of the nail is 5 cm from the contact point. what is the magnitude and direction of the force exerted by the hammer claws on the nail
1 answer
1 answer