The center of mass of the arm shown in the figure is at point A. Find the magnitudes (in N) of the tension force
Ft
and the force
Fs
which hold the arm in equilibrium. (Let 𝜃 = 23.5°.) Assume the weight of the arm is 48.7 N.
An arm is extending horizontally to the right with a cutout that shows muscles in the upper arm. A horizontal dashed line extends along the bottom of the muscles and another along the top. Point O is labeled at the left end of the arm on the lower horizontal line. An unlabeled point lies 8.00 cm horizontally to the right of point O. Point A lies 29.0 cm horizontally to the right of point O. Three forces are acting on the arm.
A force vector Fs acts down and right on point O.
A force vector Ft acts up and left on the unlabeled point at an acute angle 𝜃 above the horizontal. Vector Ft is longer than vector Fs.
A force vector Fg acts vertically down on point A. Vector Fg is shorter than vector Fs.
To find the magnitudes of the tension force Ft and the force Fs, we can use the concept of equilibrium. In equilibrium, the net force acting on an object is equal to zero.
Let's break down the forces acting on the arm:
1. The weight force Fg acting vertically downward on point A. The weight force can be calculated as Fg = m * g, where m is the mass and g is the acceleration due to gravity. In this case, the weight of the arm is given as 48.7 N.
2. The force Ft acting up and left on the unlabeled point. We need to determine the magnitude of this force.
3. The force Fs acting down and right on point O. We need to determine the magnitude of this force.
To find the magnitudes of Ft and Fs, we can set up two equations using the concept of equilibrium in both the vertical and horizontal directions.
In the vertical direction:
Sum of all vertical forces = 0
Ft * sin(theta) - Fg = 0
In the horizontal direction:
Sum of all horizontal forces = 0
Ft * cos(theta) - Fs = 0
Now, let's solve these equations:
From the vertical equilibrium equation:
Ft * sin(23.5°) - 48.7 N = 0
Ft * sin(23.5°) = 48.7 N
Ft = 48.7 N / sin(23.5°)
From the horizontal equilibrium equation:
Ft * cos(23.5°) - Fs = 0
Fs = Ft * cos(23.5°)
Let's calculate the values using the given angle 𝜃 = 23.5°:
Ft = 48.7 N / sin(23.5°) ≈ 108.64 N
Fs = Ft * cos(23.5°) ≈ 98.36 N
Therefore, the magnitudes of the tension force Ft and the force Fs are approximately 108.64 N and 98.36 N, respectively.