- #1
Brianjw
- 40
- 0
I seem to be stuck yet again on 2 different problems.
First:
A claw hammer is used to pull a nail out of a board. The nail is at an angle of 60.0 degrees to the board, and a force F1 of magnitude 500 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 F2 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 500-N force to the nail? (You can ignore the weight of the hammer.)
I started by summing the torques about point A which gave me:
[tex] \sum\tau = F_2 * .3 + .08*F_1*tan60 = I\alpha [/tex]
but, since the weight of the hammer isn't given, you're not able to find I, even if I could find alpha, I would still be an unknown. Tried summing force in the X and Y but that adds more variables.
The next one is:
A mass of 11.7 kg, fastened to the end of an aluminum wire with an unstretched length of 0.540m , is whirled in a vertical circle with a constant angular speed of 117 rev/min. The cross-sectional area of the wire is 1.50×10-2 cm^2
Calculate the elongation of the wire when the mass is at the lowest point of the path. Take the Young's Modulus of aluminum to be 7×10^10 Pa and the free fall acceleration to be 9.80 .m/s^2.
First I calculated omega, (117 Rev/ 60secs) *(2pi/1rev) = 12.25 rad/s
I need to find the radius next in order to find out the downward velocity to find the total downward force, at least I think.
I thought maybe the cross-sectiona area = pi * r^2, but not sure if that's right. Once I find the downward velocity, I was planning of finding the stress and strain to find the rest.
Anyone able to tell me if my methods are sound? and if so, what am I missing?
First:
A claw hammer is used to pull a nail out of a board. The nail is at an angle of 60.0 degrees to the board, and a force F1 of magnitude 500 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 F2 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 500-N force to the nail? (You can ignore the weight of the hammer.)
I started by summing the torques about point A which gave me:
[tex] \sum\tau = F_2 * .3 + .08*F_1*tan60 = I\alpha [/tex]
but, since the weight of the hammer isn't given, you're not able to find I, even if I could find alpha, I would still be an unknown. Tried summing force in the X and Y but that adds more variables.
The next one is:
A mass of 11.7 kg, fastened to the end of an aluminum wire with an unstretched length of 0.540m , is whirled in a vertical circle with a constant angular speed of 117 rev/min. The cross-sectional area of the wire is 1.50×10-2 cm^2
Calculate the elongation of the wire when the mass is at the lowest point of the path. Take the Young's Modulus of aluminum to be 7×10^10 Pa and the free fall acceleration to be 9.80 .m/s^2.
First I calculated omega, (117 Rev/ 60secs) *(2pi/1rev) = 12.25 rad/s
I need to find the radius next in order to find out the downward velocity to find the total downward force, at least I think.
I thought maybe the cross-sectiona area = pi * r^2, but not sure if that's right. Once I find the downward velocity, I was planning of finding the stress and strain to find the rest.
Anyone able to tell me if my methods are sound? and if so, what am I missing?