- #1

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_______ N · m counterclockwise

I just did:

τ

_{A}= [ (83.1 N)sin 48 ](1.20 cm) = 74.1

Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

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- Thread starter riseofphoenix
- Start date

- #1

- 295

- 2

_______ N · m counterclockwise

I just did:

τ

Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

- #2

- 265

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The distance is given in cm. You have to use m to get energy in J.

Also, I think the angle between the 2 vectors is 90°+48.0°

- #3

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Also, I think the angle between the 2 vectors is 90°+48.0°

Ohh ok...

thanks!

What about this one?

A cook holds a 1.13-kg carton of milk at arm's length (see the figure below). What force vector F =

I tried doing...

1) F

Fg = (1,13 kg)(9.81)

Fg = 11.08

2) T

3) T

4) Set T

3.997 = (F

133 N = F

INCORRECT: Your response differs from the correct answer by more than 10%. Double check your calculations.

- #4

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For the carton, the rotation is about the bone joint which is 33cm from the weight vector, not 25cm.

For the torque created by the muscle, draw out the

- #5

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For the torque created by the muscle, draw out therandFvectors and determine the angle between them from your drawing.

Ok this is what I did after looking it up online...

1) Determine the horizontal and vetical components of the force exerted by the biceps muscle where you see an angle - make a triangle to determine which side is cos and which side is sin.

(F

(F

2) I googled this but I dont understand the "determining axis of rotation" part...

ƩT: -F

-(1.13)(-9.81)(33 cm) - F

365.8 - F

365.8 = F

45.7 = F

176.7 N = F

I don't understand how they got the idea to add [F x r of the WHOLE ARM] MINUS (F

- #6

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When the forearm moves, it rotates about the elbow joint - this is the axis of rotation, (imagine a line through the elbow perpendicular to the diagram).

To calculate the torque, you need the force, and the distance of this force from the axis of rotation.

For the carton, the weight vector is 33cm from this axis.

For the biceps muscle, the force in the diagram is 8cm from this axis.

- #7

- 295

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To calculate the torque, you need the force, and the distance of this force from the axis of rotation.

For the carton, the weight vector is 33cm from this axis.

For the biceps muscle, the force in the diagram is 8cm from this axis.

Ohhh ok that makes perfect sense!

So whenever Im dealing with a Torque problem, I ALWAYS have to determine my main axis of rotation??

- #8

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Yes, this is the first thing you should determine because you need to have the distance from the axis of rotation to the point where the force is being applied.

- #9

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Ok thanks!!

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