Figure P8.68 shows a claw-hammer as it is being used to pull a nail out of a hor

Question

Figure P8.68 shows a claw-hammer as it is being used to pull a nail out of a horizontal board. If a force of magnitude 150 N is exerted horizontally as shown,A) Find the magnitude of the force exerted by the hammer claws on the nail. Assume that the force that the hammer exerts on the nail is parallel to the nail, and perpendicular to the position vector from the point of contact. You may assume that the hammer is in static equilibrium: the sum of the torques about any point is zero, as is the sum of the forces.
B) Find the magnitude of the force of friction exerted by the surface on the hammer.
C) Find the magnitude of the normal force exerted by the surface on the hammer.
D) Find the magnitude of the total force exerted by the surface on the hammer.
E) Find the angle that the net force exerted by the surface makes with respect to the x-axis. (Answer in degrees as deg).

Figure P8.68 shows a claw-hammer as it is being used to pull a nail out of a horizontal board. If a force of magnitude 150 N is exerted horizontally as shown, Find the magnitude of the force exerted by the hammer claws on the nail. Assume that the force that the hammer exerts on the nail is parallel to the nail, and perpendicular to the position vector from the point of contact. You may assume that the hammer is in static equilibrium: the sum of the torques about any point is zero, as is the sum of the forces. Find the magnitude of the force of friction exerted by the surface on the hammer. Find the magnitude of the normal force exerted by the surface on the hammer. Find the magnitude of the total force exerted by the surface on the hammer. Find the angle that the net force exerted by the surface makes with respect to the x-axis. (Answer in degrees as deg)

Explanation / Answer

a) The MA is (30/5) = 6.
Vertical component of force on nail = (150 x 6) = 900N.
Component at 30 degrees = (cos 30 x 900) = 779.42N.
b) Horizontal component of force on nail = (sin 30 x 900), = 450N. friction.
c) Normal force is same as vertical component, = 900N.
d) Sqrt.(900^2 + 450^2) = 1,006.23N.
e) Angle = arctan (450/900) = 26.6 degrees from vertical.

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