A 50.0 kg grindstone is a solid disk 0.510 m in diameter. You press an ax down on the rim with a normal force of F = = 140 N (Figure 1). The coefficient of kinetic friction Part A between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N·m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s? Express your answer with the appropriate units. HÁ Figure 1 of 1 F1 = 100.2 N m = 50.0 kg Submit Previous Answers Request Answer F Incorrect; Try Again; 5 attempts remaining Part B After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min? Express your answer with the appropriate units.

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A 50.0 kgkg grindstone is a solid disk 0.510 mm in diameter. You press an ax down on the rim with a normal force of FF = 140 NN (Figure 1). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N⋅m6.50N⋅m between the axle of the stone and its bearings.

A 50.0 kg grindstone is a solid disk 0.510 m in diameter.
You press an ax down on the rim with a normal force of
F =
= 140 N (Figure 1). The coefficient of kinetic friction
Part A
between the blade and the stone is 0.60, and there is a
constant friction torque of 6.50N·m between the axle
of the stone and its bearings.
How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest
to 120 rev/min in 9.00 s?
Express your answer with the appropriate units.
HÁ
Figure
1 of 1
F1 = 100.2
N
m = 50.0 kg
Submit
Previous Answers Request Answer
F
Incorrect; Try Again; 5 attempts remaining
Part B
After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is
needed to maintain a constant angular speed of 120 rev/min?
Express your answer with the appropriate units.
Transcribed Image Text:A 50.0 kg grindstone is a solid disk 0.510 m in diameter. You press an ax down on the rim with a normal force of F = = 140 N (Figure 1). The coefficient of kinetic friction Part A between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N·m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s? Express your answer with the appropriate units. HÁ Figure 1 of 1 F1 = 100.2 N m = 50.0 kg Submit Previous Answers Request Answer F Incorrect; Try Again; 5 attempts remaining Part B After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min? Express your answer with the appropriate units.
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