Question

1)A wheel is used to turn a valve stem on a water valve. If the wheel radius is 1 foot and the stem, (axle), radius is .5 inches, what is the mechanical advantage of the wheel and axle? 2)How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1? 3)What is the linear distance traveled when a 2.5' diameter wheel makes one revolution? 4)A compressor motor has a 1.00 inch output shaft (axle) that drives a 4.00inch pulley (wheel). If the motor output shaft supplies 24 lbs of force, ideally how much force does the pulley apply to the drive belt wrapped around it? 5) If the belt pulley's force is measured to be 4 lbs, what is the AMA wheel and axle here?6) What is the efficiency of the compressor motor – belt pulley wheel and axle?

Answer 1

Answer:

1.) 2.4

2.) 112 lbs

3.) 7.85 inches

4.) 6 lbs

5.) 2 lbs

6.) 67%

Explanation:

Given that

Radius of the wheel R = 1 foot

1 foot = 12 inches.

Radius of the axle r = 5 inches

1.) The mechanical advantage MA is :

MA = R/r = 12/5 = 2.4

2.) How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1?

MA = Load / effort

Where effort = 80 lbs

Substitute MA and effort into the formula

2.4 = Load / 80

Cross multiply

Load = 2.4 × 80 = 192 lbs

The resistance force to be overcome will be

Force = load - effort

Resistance force = 192 - 80 = 112 lbs

3) What is the linear distance traveled when a 2.5' diameter wheel makes one revolution

One revolution = 2π

Radius = 2.5 /2 = 1.25 inches

Linear distance S = angular distance Ø × radius r

S = Ør

S = 2π × 1.25

S = 7.85 inches

4. ) given that

Wheel radius R = 4

Axle radius r = 1

MA = 4/1 = 4

MA = Load / effort

4 = 24/ effort

Effort = 24/4 = 6 lbs

5.) 6 - 4 = 2lb

6.) Efficiency = MA / VR × 100

Efficiency = 4 / 6 × 100

Efficiency = 67%