A claw hammer is used to pull a nail out of a board (?Fig. P11.48?). The nail is at an angle of 60o to the board, and a force of magnitude 400 N applied to the nail is required to pull it from the board. The hammer head contacts the board at point A, which is 0.080 m from where the nail enters the board. A horizontal force is applied to the hammer handle at a distance of 0.300 m above the board. What magnitude of force is required to apply the required 400-N force (F1) to the nail? (Ignore the weight of the hammer.)

Solution 52P Step 1: Suppose, “r” is the lever arm of the force F’ which is acting opposite to the F . The force 1 1 F’1 is exerted by the nail which is in the opposite direction of the force exerted by the hammer. We know that, F ,F 1d2’ are t1 forces acting on the x-y plane So, we can write, the total moment = 0 That is, Step 2: The total moment of the system, h is the vertical height where force F is acting and it is provided. h = 0.300 m 2 We should find out the perpendicular distance of force F’ . 1 We can represent it by using the diagram We can redraw the point of contact of the hammer as this. The triangle is a right angled triangle. So, we have have to find out the perpendicular distance “r”. So, we can write, sin 60 = r / 0.08 m Or, r = 0.08 m × 0.8660 = 0.0693 m