Calculating Applied Force for Lifting a 450 kg Car with a Scissors Jack

[Chemlach]

1. The Problem is a simple scissors jack has lifted a 450 kg car and the angle of the arms of the jack are 15° as seen in the attached drawing



2. The question is what is the applied force required to lift the 450 kg mass?



3. I drew my free body diagram and got that F=mg/2cos15°, which comes out to 2280. But the answer is 1180 so I must have missed something. Could you let me know where I went wrong?

 

[gneill]

1. The Problem is a simple scissors jack has lifted a 450 kg car and the angle of the arms of the jack are 15° as seen in the attached drawing



2. The question is what is the applied force required to lift the 450 kg mass?



3. I drew my free body diagram and got that F=mg/2cos15°, which comes out to 2280. But the answer is 1180 so I must have missed something. Could you let me know where I went wrong?

The weight of the car will be split into two forces traveling down the jack struts from the top. By symmetry these forces will be the same, and their resultant will equal the weight of the car.

Now consider the joint where one of the struts meets one of the bottom struts. The force transmitted down the upper strut is met with a similar force coming up from the bottom strut (again by symmetry) directed along the bottom strut. The force required to 'restrain' the net outward force (resultant of those two forces) is what you're looking for.