A string that passes over a pulley has a 0.351kg mass attached to one end and a
ID: 1379478 • Letter: A
Question
A string that passes over a pulley has a 0.351kg mass attached to one end and a 0.655kg mass attached to the other end. The pulley, which is a disk of radius 9.00cm , has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? A hand-held shopping basket 57.0cm long has a 1.76kg carton of milk at one end, and a 0.672kg box of cereal at the other end. Where should a 1.85kg container of orange juice be placed so that the basket balances at its center? A cat walks along a uniform plank that is 4.00m long and has a mass of 6.00kg (Figure 1). The plank is supported by two sawhorses, one 0.440m from the left end of the board and the other 1.50m from its riaht end. If the cat has a mass of 2.8kg , how close to the right end of the two-by-four can it walk before the board begins to tip? Express your answer using two significant figures.Explanation / Answer
Question 1)
The frictional torque must be the difference between the mass torques...
That is FL - FL
(.655)(9.8)(.09) - (.351)(9.8)(.09) = .268 Nm
Question 2)
The center of the basekt is at 28.5 cm. The orange juice needs to be toward the side of the lighter object, so by torques...
(1.76)(9.8)(.285) = (1.85)(9.8)(x) + (.672)(9.8)(.285)
x = .168 m from the center
That is 16.8 cm from the center
Problem 3)
Placing the pivot point of the right support will solve by torques.
The left support will apply zero force or torque right at the balance point, so...
Fr = Fr
The center of the board is at 2 meters. Since the right support is at 1.5 meters from the end, the center is .5 meters from the support
(6)(9.8)(.5) = (2.8)(9.8)(x)
x = 1.07 m from the support
Subtract from 1.5 and get .43 m from the right edge
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