the right side of the fulcrum and a 0.050-m-long part on the left side supports the back of the car. How hard must you push down on the handle so that the lever exerts an 9000-N force to lift the back of the car?
9000/(1.6/0.050) 281.25N.
Problem. As the short end of the lever is only 5 cm long you can only lift the car by this amount approximately. And when you do, the tyres of the car are still on the ground. You have extended the suspension slightly but not lifted the car from the ground. If you can get the lever under the axle instead then a) you left the wheel from the road and b) you halve the force needed. In this way you can move the fulcrum back increasing the amount of lift while still keeping the force at acceptable levels. By the way, once you HAVE lifted the car how are you going to change the wheel while you are still at the other end of this 1.6 m lever. So although the answers given are true enough, practical experience shows that the PROCESS as written cannot work.
This is a torque problem. When you push the right side of the lever down, you are producing clockwise torque Torque F * d, d 1.6 Torque F * 1.6 Counter clockwise torque 9000 * 0.05 450 F * 1.6 450 F 450/1.6 281.25 N I hope this helps you to understand how to solve this type of problem.
