Question

nutaud 4) Read the following and revise your free body diagram as necessary. A common confusion when drawing extended free body diagrams is how to draw the force of gravity. For a static equilibrium situation like this, you should draw a single arrow to represent the force of gravity on an extended object. The tail of this vector is placed at the center of mass of the object. It is good practice to pause at this point and think about whether the directions of the forces you drew are consistent with static equilibrium. Thinking of the shoulder joint as the axis of rotation, in what sense (CW, CCW, or zero) are the torques due to each of the forces you drew? Is it possible for all the torques to balance out? What if you thought of the center of mass of the arm as the axis of rotation? Will the forces you drew enable Fet x and Fnet y to be zero? Suppose your "knowns" include: The mass of the arm (m) The distance from the shoulder joint to the arm's center of mass (D) to clai The distance from the shoulder joint to the deltoid's attachment point (di WO 2sot The angle between the deltoid and the arm (0) SAYE Your "unknowns" include: The magnitude of the force on the arm by the deltoid (F) . s1ot The x and y components of the force on the arm by the shoulder joint (N and Ny) 5) Choosing the shoulder joint as the axis of rotation, fill in the static equilibrium condition for torques: T = 0 with the details of this problem and then solve for F. Your answer should be an expression in terms of the "knowns". to hio srd iot : worda ) ms avds diw snil lanisd ss tolblo oino we C (E eturnio y lws 195luode 4.0 kg, D= 38 cm, d 17 cm, and 0 15° to find a value for F. 6) Plug in the values m

Answer 1

This is how the forces are present:

F sin Center of mass Shoulder joint X F cos d mg

So, writing down the force equations, knowing that the shoulder joint is a ball and socket joint and restricts translation in all three directions but allows rotation about all three axes, we have:

Force balance:

, along the vertical direction

NyFsin(0) = mg

and

, along the horizontal direction.

N Fcos(0)

Torque balance: (assume CCW positive)

Taking all torques about the shoulder joint, we get:

Fsin (0)d mgD 0 Fdsin (0)mgD mgD F dsin(0)

This the the equation for the force by the deltoid in terms of the knowns.

Substituting the values given we get: F =338.89 N