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r T T r w w = 0 .

The magnitude of tension in the muscle is obtained by solving [link] :

T = r w r T w = 13.0 1.5 (50 lb) = 433 1 3 lb 433.3 lb.

The force at the elbow is obtained by solving [link] :

F = w T = 50.0 lb 433.3 lb = −383.3 lb.

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

F = 383.3 lb = 383.3 ( 4.448 N ) = 1705 N downward T = 433.3 lb = 433.3 ( 4.448 N ) = 1927 N upward.

Significance

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Solution

Suppose we adopt a reference frame with the direction of the y -axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y -components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in [link] , indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles θ T and θ w that the forces T M and w (respectively) make with their lever arms. In the definition of torque given by [link] , the angle θ T is the direction angle of the vector T M , counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle θ w is measured counterclockwise from the radial direction of the lever arm to the vector w . Done this way, the non-zero torques are most easily computed by directly substituting into [link] as follows:

τ T = r T T sin θ T = r T T sin β = r T T sin 60 ° = + r T T 3 / 2 τ w = r w w sin θ w = r w w sin ( β + 180 ° ) = r w w sin β = r w w 3 / 2 .
Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force Tm is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Force Tm forms and angle theta tau that is equal to beta with the direction of the lever arm. Force W forms an angle theta w that is equal to the sum of beta and Pi with the direction of the lever arm.
Free-body diagram for the forearm for the equivalent solution. The pivot is located at point E (elbow).

The second equilibrium condition, τ T + τ w = 0 , can be now written as

r T T 3 / 2 r w w 3 / 2 = 0 .

From the free-body diagram, the first equilibrium condition (for forces) is

F + T w = 0 .

[link] is identical to [link] and gives the result T = 433.3 lb . [link] gives

F = T w = 433.3 lb 50.0 lb = 383.3 lb.

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Check Your Understanding Repeat [link] assuming that the forearm is an object of uniform density that weighs 8.896 N.

T = 1963 N ; F = 1732 N

Got questions? Get instant answers now!

A ladder resting against a wall

A uniform ladder is L = 5.0 m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in [link] . The inclination angle between the ladder and the rough floor is β = 53 ° . Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction μ s at the interface of the ladder with the floor that prevents the ladder from slipping.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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