Concept explainers
A 215-kg robotic arm at an assembly plant is extended horizontally (Fig. P14.32). The massless support rope attached at point B makes an angle of 15.0° with the horizontal, and the center of mass of the arm is at point C. a. What is the tension in the support rope? b. What are the magnitude and direction of the force exerted by the hinge A on the robotic arm to keep the arm in the horizontal position?
FIGURE P14.32
(a)
Tension in the support rope.
Answer to Problem 32PQ
The tension in the support rope is
Explanation of Solution
The figure given below is the free body diagram of the robotic arm.
In the equilibrium state of the system, the net torque experienced by the robotic arm is zero. Take shoulder joint at point A as the pivot point. The forces acting on the body are the force of tension on the rope and the force at robotic arm due to gravitational field. Torque acts on the arm due to these forces.
Write the equation to find the torque produced by the tension on the rope.
Here,
Write the equation to find the torque acting on the arm due to earth’s gravity.
Here,
Write the equation to find the gravitational force.
Substitute the above equation in (II) to get
The force of gravity is in down ward direction, therefore the torque due to gravitational force has a negative sign.
Add equations (III) and (I) to find the sum of torques at the arm.
Since the sum of torque is zero in equilibrium condition, equate equation (IV) to zero and solve for
Conclusion:
Substitute
Therefore, the tension in the support rope is
(b)
The magnitude and direction of force exerted by hinge
Answer to Problem 32PQ
The magnitude of the force exerted by hinge
Explanation of Solution
At equilibrium the sum of both x and y components of all the force is equal to zero.
The forces in the y direction are the force exerted by the hinge, force due to the tension on rope, and the force due to gravity.
Write the equation to find the y component of force due tension on the rope.
Here,
Write the equation to find the force due to gravity.
Add equation (VI) and (VII) to get the sum of forces.
In addition to these two forces there is the force exerted by the hinge on the arm of the robot. Let this force be
Add
Here,
At equilibrium, the sum of y component of forces is equal to zero. Equate equation (VIII) to zero and solve for
Substitute equation (V) in (IX) and solve for
The forces acting in the x direction are the force due to tension on rope and the horizontal component of force exerted by the hinge.
Write the equation to find the x component of the tension force on the rope.
Here,
Substitute equation (V) in (XI).
Write the equation to find the sum of forces in the x direction.
Here,
Substitute equation (XII) to the above equation to get
In equilibrium, sum of forces in the x direction is zero. Equate equation (XIII) to zero.
Solve equation (XIV) to get
Write the equation to find the resultant force exerted by hinge.
Here,
Write the equation to find the angle made by the resultant force with the horizontal.
Here,
Conclusion:
Substitute
Substitute
Substitute
Substitute
Therefore, the magnitude of the force exerted by hinge on arm is
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Chapter 14 Solutions
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