I have now made some calculations to solve both of the parts of this question.
a) Anti clockwise: F x 0.5
Clockwise: Y x 2.5
Anti clockwise = Clockwise
0.5 x F = 2.5 x Y
0.5 x 18 = 2.5 x Y
Forces against gravity: X + Y
Gravity forces: Fg + F
Forces against gravity = Gravity forces
X + Y =...
It is because our teacher didn't go through any of this with us and I have no clue what I am doing. I don't even know what and arbitrary point is or how to pick a point :(
The new equation will be:
X - 30N x 1m = 30Nm
18N x 0.5m = 9Nm
Y - 30N x 0.5m = 15Nm
18N x 36Nm
X = 39Nm
Y = 51Nm
Somehow it does not add up that there is more force applied on support Y, that is the furthest away from the extra weight...
So how would that work on support Y? Is it 2m away? Then I have two separate weights with different distances to the two supports, so it will be two separate equations?
There is a weight of 18N on the left hand end of the plank, which I added on after calculating the force of the plank only on the supports.
Here is the full first question: A uniform plank, of weight 30N and length 3m, resting on two supports. The supports are 0.5m and 2.0m from the left hand...
The length from the center of the plank to support X is 1m and to Y is 0.5m, so the force at point X multiplied by the length should be equal to the force multiplied by the length at point Y
I added my calculations from the first part of the question, but need help in understanding the second part. I have been searching my books and the internet and can't find a formula to calculate the second part of this question.
Homework Statement
State an expression for the vertical force at the elbow joint FE in terms of the other forces acting when the arm is held horizontal with the dumbbell in the hand.
Homework Equations
No equations needed.
The Attempt at a Solution
Since muscles can not expand beyond...