A band saw uses a toothed band to cut material. The object being cut is shown...

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A band saw uses a toothed band to cut material. The object being cut is shown in dashed lines in the left of the drawing, and requires 100 N to push the blade's teeth through the object. The coefficient of static friction between the belt and drive pulley is 0.5. Assume the band saw loop is on the verge of slipping relative to the drive pulley. a) Compute the tension in the upward moving side of the belt, after the drive pulley b) The torque that must be applied to the drive pulley (the driven and drive pulleys are the same diameter) c) The torque that the motor must apply The band saw loop must be pre-tensioned by adjusting the driven pulley upwards while the saw is off. But even when the saw is on and cutting, the sum of the tensions in the upward and downward moving segments of the loop is equal to the sum of the tensions from the pre-tensioning adjustment. Therefore, given that the total tension remains constant: d) what is the minimum upward force that must be applied at the center of the driven pulley in order to cut through the object given at the beginning of the problem? Drive pulley D=400 mm 1 -Driven pulley D=500 mm Saw band Frame -V-belt -Electric motor dm 125 mm Motor base A band saw uses a toothed band to cut material. The object being cut is shown in dashed lines in the left of the drawing, and requires 100 N to push the blade's teeth through the object. The coefficient of static friction between the belt and drive pulley is 0.5. Assume the band saw loop is on the verge of slipping relative to the drive pulley. a) Compute the tension in the upward moving side of the belt, after the drive pulley b) The torque that must be applied to the drive pulley (the driven and drive pulleys are the same diameter) c) The torque that the motor must apply The band saw loop must be pre-tensioned by adjusting the driven pulley upwards while the saw is off. But even when the saw is on and cutting, the sum of the tensions in the upward and downward moving segments of the loop is equal to the sum of the tensions from the pre-tensioning adjustment. Therefore, given that the total tension remains constant: d) what is the minimum upward force that must be applied at the center of the driven pulley in order to cut through the object given at the beginning of the problem? Drive pulley D=400 mm 1 -Driven pulley D=500 mm Saw band Frame -V-belt -Electric motor dm 125 mm Motor base