calculation max load on a scissor jack lifting a car


by vpatel28
Tags: engineering, mechanical, physics, scissor jack
vpatel28
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#1
Nov7-09, 10:24 PM
P: 5
1. The problem statement, all variables and given/known data

I have to determine the max load a scissor jack must support when lifting a 6000lb car. I know where the center of gravity is acting and the potential locations (distances from the center of gravity) the car will be jacked up. I know the max force is less than 6000 lb.

2. Relevant equations



3. The attempt at a solution

I tried taking the moments about one of the tires and setting them equal to zero to solve for the max force at each location, but I know I'm doing something wrong.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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tiny-tim
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#2
Nov8-09, 06:41 AM
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Hi vpatel28! Welcome to PF!
Quote Quote by vpatel28 View Post
I tried taking the moments about one of the tires and setting them equal to zero to solve for the max force at each location, but I know I'm doing something wrong.
That should work
show us what you got!
vpatel28
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#3
Nov8-09, 10:47 AM
P: 5
I've attached a diagram of the geometry of the car. The more I think about the problem, I feel like I need to know which two tires the car will lean on for each case. Inspection tells me that it will lean on the two side tires for both cases.

Do I want to take the moment about one of the tires that car is leaning on? If I do that I get the following:

Case 1 - Lean on Back wheels
(6000)(8 ft) + -F(10) = 0
F= 4800 lb

Case 2 - Lean on Side Wheels
(-6000)(3) + (F)(6) = 0
F = 3000 lb

I'd appreciate any help in directing me in the right direction

tiny-tim
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Nov8-09, 11:40 AM
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calculation max load on a scissor jack lifting a car


Quote Quote by vpatel28 View Post
I've attached a diagram of the geometry of the car.
erm noooo!
Do I want to take the moment about one of the tires that car is leaning on? If I do that I get the following:

Case 1 - Lean on Back wheels
(6000)(8 ft) + -F(10) = 0
F= 4800 lb

Case 2 - Lean on Side Wheels
(-6000)(3) + (F)(6) = 0
F = 3000 lb
Difficult to be sure without seeing a diagram,
but that certainly looks ok!
vpatel28
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#5
Nov8-09, 12:16 PM
P: 5
Sorry about that. Here is the attachment with the diagram
Attached Files
File Type: pdf Scissor Jack.pdf (59.3 KB, 188 views)
tiny-tim
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Nov8-09, 02:19 PM
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Quote Quote by vpatel28 View Post
I've attached a diagram of the geometry of the car. The more I think about the problem, I feel like I need to know which two tires the car will lean on for each case. Inspection tells me that it will lean on the two side tires for both cases.
ok, from the diagram the wheels are 12 ft apart front-to-back, and 6 ft apart side-to-side.

The jack points are 2ft behind or in front of the nearest wheel.

The c.o.m. is 4ft behind the front wheels and 8ft in front of the back wheels.
Case 1 - Lean on Back wheels
(6000)(8 ft) + -F(10) = 0
F= 4800 lb

Case 2 - Lean on Side Wheels
(-6000)(3) + (F)(6) = 0
F = 3000 lb
Yes, those equations are fine, except that in case 1, if the c.o.m. is nearer the front, won't the car overbalance onto the side?
vpatel28
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#7
Nov8-09, 02:25 PM
P: 5
If the the car leans on the side wheels for both cases then isn't the max force the same in both cases?
tiny-tim
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Nov9-09, 02:36 AM
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Quote Quote by vpatel28 View Post
If the the car leans on the side wheels for both cases
It doesn't it leans on the front wheels for a back jacking point, and on the side wheels for a front jacking point

can you prove that?


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