# Homework Help: Torque Question

1. Dec 5, 2005

### SS2006

http://img210.imageshack.us/img210/3923/q17tp.jpg [Broken]
Question 3.12 (ignore the 0.80 i have written under it)

i dont know what to do, well first i started the torque fbd at the first tire

torque 1) - 700 * 0.25L (negative cause its clockwise)
torque 2) - 8000 *0.5L
torque3) - 900 *0.75L
torque4) F * 1L

-175 - 4000 - 675 + F
4850 = F
as you see thats not the right answer
i know i did something wrong let me know

Last edited by a moderator: May 2, 2017
2. Dec 5, 2005

### dicerandom

Looks like you need to re-examine the distances that you're using in your torques. Re-read the problem and take a closer look at the figure. Also remember that you'll need to divide your final answer by 2 since there are two wheels at the front and back :)

3. Dec 5, 2005

### SS2006

i knew i had the distances werong
im blanked for some reason :(

4. Dec 5, 2005

### SS2006

from that same question sheet btw

3.16
i got 2.9kN as teh tension in the tie rope, but whaen they ask for force of the hinge on the beam, believe it or not i dotn remember being taught this, what do you do in this case.

as for 3.15
they ask for the force components at thehinges
once again i did get 670N, but in the answers theres H=670N V= 1.6kN
i really dont know this, a quick lesson would be apprecaited thans

5. Dec 5, 2005

### dicerandom

There's a basic strategy for attacking these kinds of problems. They're generally called statics, by the way, since the situations are supposed to be static (i.e. nothing's moving).

Basically what you want to do is to determine exactly what forces and torques are acting on the object in question. That would be the floor of the car in your first question and the beams in your second two.

Once you've determined the forces that are acting on the object break them up into components and use Newton's second law (F=ma) to get a system of equations relating the various forces to one another. In these cases nothing's moving so your acceleration will always be zero, i.e. you'll be setting the sums of your forces and torques equal to zero.

For instance, in the case of the situation in Figure 3-15 I'll go through the forces that are acting on the beam in the vertical direction. You have a gravitational force from the weight of the beam itself that's pulling down, you have the weight of the mass hanging off the end which is also pulling down, you have the vertical component of the force exerted by the rope which is pulling up, and you have a vertical component of force from the hinge which is pushing down. The sum of all these forces must equal zero.

Simmilarly you should be able to come up with equations relating the horizontal components of force which are acting on the beam, and all of the torques as well. Once you've done this you will have a system of three equations with three unknowns that you should be able to solve using standard algebraic techniques.

Last edited: Dec 5, 2005
6. Dec 5, 2005

### SS2006

tha tmade alot fsense thanks so much
i understood vetical part very weel
but what torques are in the horizontal part?