A log weighing 800 lb is lifted by a pair of tongs as shown. Determine the forces exerted at E a

A log weighing 800 lb is lifted by a pair of tongs as shown....

Key Concepts

Free-Body Diagram

A free-body diagram is a simplified representation of an object, isolating it to show all external forces and moments acting upon it. This tool is fundamental in statics for visualizing the problem, setting up equilibrium equations, and systematically solving for unknown forces such as those at the points of contact in the lifting mechanism.

Static Equilibrium

This concept refers to a state where a body remains at rest, meaning that the sum of all forces and the sum of all moments (or torques) acting on the body are equal to zero. Solving problems involving static equilibrium requires setting up equations that balance forces and moments, ensuring that the object does not accelerate or rotate.

Moment (Torque) Calculation

Moment or torque is the rotational effect produced by a force acting at a distance from a pivot point. In the analysis of lifting objects with tongs, calculating moments is essential to determine the forces at different points by considering how the applied force creates a turning effect about a fixed axis.

Lever Principle

The lever principle explains how a force applied at one point can be amplified or reduced at another point by using a rigid bar rotating about a fulcrum. This principle is crucial in understanding how the magnitudes of forces at different points along a tong, or similar devices, are related based on the distances from the pivot or fulcrum.

Transcript

00:01 We need to find the forces exerted at the point e and the point of whenever this log of 800 pounds is lifted up using the tons.

00:11 So i've drew out this free body diagram, and the first thing we're going to do is find the angle theta in the free body diagram.

00:20 And when solving this problem, we're going to assume that ac, the links from ac and links from bd are on their own axis.

00:31 So to find angle theta, we can write that as...

00:36 Tangent inverse of 2 .5 divided by 3 and this will give us an angle of 39 .805 degrees.

00:51 And now we're going to say that the sum of forces about the wide direction is equal to 0.

01:00 And we're going to solve for the forces at point a and point b.

01:06 So we can write that as 800 minus a of y minus a of b, a of b, a of f.

01:18 Sorry so 800 minus a of y minus b of y is equal to zero which we can also write as a of y plus b of y is equal to 800 so now we're going to take the moments about the about the point b so the sum of moments about point b is equal to zero and we can write that as a of y times three minus 800 times 1.

02:08 5 is equal to 0.

02:12 And solving this expression for a of y, we'll get a force value of 400 pounds.

02:27 And we can now solve for the magnitude of the force at a by saying that the magnitude of force a is equal to, so we write that a of y is equal to the magnitude of the force a times the sine of data.

02:53 So we take this equation, soft ray and plug in the known values.

03:01 So we're going to have that a is equal to a of y.

03:11 So a of y divided by a sine of 39 .805.

03:25 And this will give us a force of 624 .819 pounds.

03:35 We can also offer the x component of the force by saying that a of x is equal to the magnitude of a times the cosine of theta.

03:47 So we plug in a and we plug in cosine of theta.

03:57 So plug in a and then times the cosine of 39 .805 degrees.

04:08 And this will give us a force of 480 pounds.

04:12 So we go back to this equation, so we can use this equation now and solve for the y component of the forks at point b.

04:31 So we're going to have 400 plus b of y is equal to 800.

04:42 And solving this for b of y, we'll get a force that is equal to 400 pounds.

04:53 And we can say that since this system is, since this system is, is symmetrical...

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