Rotational dynamics and static equilibrium question

[hkstyles]

Please help with this question. I don't know how to approach this difficult question

a crossing guard holds a STOP sign at arm's length, as shown. Her arm is horizontal, and we assume that the deltoid muscle is the only muscle supporting her arm. The wright of her upper arm is Wu = 18 N, the weight of her lower arm is Wl = 11 N, the weight of her hand is Wh = 4.0N, and the weight of the sign is Ws = 8.9 N. The location where each of these forces acts on the arm is indicated in the figure. A force of magnitude fd is exerted on the humerus by the deltoid, and the shoulder joint exerts a foce on the humerus with horizontal and vertical components given by fx and fy , respectively. (a) is the magnitude of fd greater than, less than, or equal to the magnitude of fx ? why?. Find (b) fd (c) fx and (d) fy

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[learningphysics]

Think about the 3 equations involved...

sum of torques about any point = 0 (1)

sum of forces in x-direction = 0 (2)

sum of forces in y-direction = 0 (3)

equation (2) should let you answer a).

equation (1) gives you part b). try to go on from there to get the other parts...

 

[ogb p]

I would approach this question by first understanding the concepts of rotational dynamics and static equilibrium. Rotational dynamics deals with the motion of objects that rotate around a fixed axis, while static equilibrium refers to the state in which an object is at rest and all forces acting on it are balanced.

In this scenario, the crossing guard's arm is in static equilibrium, meaning that the sum of all the forces acting on her arm is equal to zero. This is because her arm is not moving and there is no acceleration.

To answer the first part of the question, we need to compare the magnitudes of fd and fx. Since fd is the force exerted by the deltoid muscle and fx is the horizontal component of the force exerted by the shoulder joint, we can conclude that the magnitude of fd is equal to or greater than the magnitude of fx. This is because the deltoid muscle is the only muscle supporting her arm, so it must exert a force at least equal to the horizontal component of the force exerted by the shoulder joint in order to keep her arm in static equilibrium.

To find the magnitude of fd, we can use the formula for torque, which is given by T = r x F, where r is the distance from the point of rotation to the point where the force is applied, and F is the force applied. In this case, the point of rotation is the shoulder joint, and the force applied is fd. So, we can write the equation as T = fd x L, where L is the length of the arm. Since the arm is held at arm's length, L = 0.9 m. We also know that the weight of the upper arm, lower arm, hand, and sign are acting at a distance of 0.3 m, 0.6 m, 0.85 m, and 0.95 m from the shoulder joint, respectively. So, we can write the equation as:

0.3 x Wu + 0.6 x Wl + 0.85 x Wh + 0.95 x Ws = fd x 0.9

Solving for fd, we get fd = 24.4 N.

To find fx and fy, we can use the equations for static equilibrium, which state that the sum of all forces in the x-direction and y-direction must be equal to zero. So, we can write the equations as:

fx + fd = 0