Find the torque about the knee

[bmandrade]

Homework Statement

I need to solve a problem where a 7.5 Kg object is attached to an ankle and leg lifts are done. There is a picture that shows the leg being at rest where the angle is 0, then legs its lift to 30 degree angle, 60 and 90 degree angle

I am suppose to find the torque about the knee due to this weight for the 4 positions (each angle)

Homework Equations

torque = rF sin (angle)

The Attempt at a Solution

So i used the equaiton above to solve this problem which gave me

at 0 degrees = 0 N.m
at 30 degrees = 15 N.m
at 60 degrees = 25 N.m
at 90 degrees = 30 N.m

is this right?

 

[LowlyPion]

Homework Statement

I need to solve a problem where a 7.5 Kg object is attached to an ankle and leg lifts are done. There is a picture that shows the leg being at rest where the angle is 0, then legs its lift to 30 degree angle, 60 and 90 degree angle

I am suppose to find the torque about the knee due to this weight for the 4 positions (each angle)

Homework Equations

torque = rF sin (angle)

The Attempt at a Solution

So i used the equaiton above to solve this problem which gave me

at 0 degrees = 0 N.m
at 30 degrees = 15 N.m
at 60 degrees = 25 N.m
at 90 degrees = 30 N.m

is this right?

How many Newtons is 7.5 kg ?

Do you have access to any Sine tables to look up the values of Sine for those angles?

How long is it again from the ankle to the knee? I missed seeing that in your statement.

 

[bmandrade]

ok 7.5 kg = 73.5 N
also the lenth of the leg is 40 cm so .4 m

 

[LowlyPion]

ok 7.5 kg = 73.5 N
also the lenth of the leg is 40 cm so .4 m

And what is that as torque over that distance - expressed as N-m?

 

[bmandrade]

.4 m x 73.5 N = 29.4

 

[LowlyPion]

.4 m x 73.5 N = 29.4

OK, so is that one of your answers and if so for what angle?

 

[bmandrade]

it would be for angle 0 right??

 

[LowlyPion]

it would be for angle 0 right??

If the leg is laying flat when just lifted then yes. 0 degrees relative to x-axis. 1 down.

Now what is the distance of the weight from the pivot when the leg is at 30 degrees?

Which trig function gives you that distance the force is acting straight down through to the pivot?

 

[bmandrade]

cos
I will have to use the .4 m as the side of the triagle the 30 degree angle makes this way i can find the distance of the 30 degree angle by using cosine

so

.4 (cos 30) = .346 m

using this i have to plug into the torque equation do i have to use sin of 30 or no because since the force is not exactly being applied at an angle right??

 

[LowlyPion]

cos
I will have to use the .4 m as the side of the triagle the 30 degree angle makes this way i can find the distance of the 30 degree angle by using cosine

so

.4 (cos 30) = .346 m

using this i have to plug into the torque equation do i have to use sin of 30 or no because since the force is not exactly being applied at an angle right??

The Cos already accounts for the angle.

Using .346 m is correct.

 

[bmandrade]

oh ok in that cas
.346 m (73.5) = 25.4 N-m for angle 30

for angle 60 .173 m (73.5) = 12.7 N-m

for angle 90 will it be 0 since cos of 90 = 0?

 

[LowlyPion]

oh ok in that cas
.346 m (73.5) = 25.4 N-m for angle 30

for angle 60 .173 m (73.5) = 12.7 N-m

for angle 90 will it be 0 since cos of 90 = 0?

Not quite. It's Cosine taken over the original .4m not the .346 at 30 degrees.

 

[bmandrade]

so i would have to use cos of 60 times .4

 

[LowlyPion]

so i would have to use cos of 60 times .4

Correct.

 

[bmandrade]

oh ok but still for the angle of 90 its 0 or no?

 

[LowlyPion]

oh ok but still for the angle of 90 its 0 or no?

Cos 90 is zero.

Multiplying by zero covers up a lot of sins.

 

[bmandrade]

so at 90 there is no torque

 

[LowlyPion]

so at 90 there is no torque

All of the weight is acting directly through the pivot when it's straight up isn't it? Hence no moment arm no torque.