Find the tension T in each cable and the magnitude and direction of the force exerted on the str

step 1 In this problem, we're going to talk about static equilibrium. So let's suppose we have a struts

Find the tension T in each cable and the magnitude and...

Key Concepts

Tension Analysis in Cables

Tension analysis in cables involves determining the force transmitted along a cable when it is subjected to loads. In scenarios with multiple cables and supports, resolving the tension forces correctly, often by decomposing them into components, is essential to ensure that the structure remains in equilibrium under the applied loads.

Reaction Forces at Pivots

Reaction forces at pivots are forces exerted by a support or hinge that restricts the movement of a body. These forces, typically having both horizontal and vertical components, must be included in the equilibrium equations. Understanding how to calculate these forces is crucial when analyzing structures with pivot points.

Static Equilibrium

Static equilibrium occurs when an object is at rest and the net force and net moment acting on it are zero. In problems like these involving cables, struts, and pivots, applying the conditions for static equilibrium helps determine unknown forces and tensions by ensuring that the sums of horizontal forces, vertical forces, and moments about any point add to zero.

Free-Body Diagram

A free-body diagram (FBD) is a simplified representation of an object and all the external forces acting on it. In problems involving static equilibrium, drawing an accurate FBD is essential to isolate the object and identify forces such as weights, tensions, and reaction forces which then allows for the correct application of equilibrium equations.

Moment Equilibrium

Moment equilibrium involves taking moments (or torques) about a chosen point and setting the sum equal to zero. This is particularly useful in structures where forces act at different points, as it provides a way to solve for unknown tensions or forces by eliminating reaction forces that do not contribute to the moment at that point.

Transcript

00:01 In this problem, we're going to talk about static equilibrium.

00:03 So let's suppose we have a struts and a block that's supported by this strut, and the strut itself is supported by a rope and a pivot.

00:18 Okay, and here we have a wall.

00:22 This is pretty much the setup of our problem in our exercise.

00:27 Let me give you a brief overview of how we generally solve these types of problems.

00:34 Let's say that this angle here is theta.

00:38 So for an object to be in static equilibrium, we need that the sum of the forces over the object be equal to 0.

00:50 That must be true.

00:51 And also the sum of the torques that act on the object must be equal to 0.

00:57 The first condition guarantees that the center mass of the object won't move.

01:02 And the second condition guarantees that the torque, i'm sorry, that the object will not rotate around the pivot.

01:13 Okay, so now we can go on to our problem.

01:16 And before going on, we just need to remember that f, actually, that the torque of tau is equal to f cross r, where r is the distance between the point where the force is being applied.

01:34 And the center of rotation.

01:39 And so the magnitude of the torque can be written as f times r times the sign of theta, where theta is the four, the angle between the force being applied and the line that connects the center of rotation to the point where the force is being applied.

01:58 Okay.

02:01 So now we can move on to our exercises.

02:04 What we have are two setups.

02:08 And in the first one i'm going to call it a we have the struts exactly as before so actually i'm not even going to use the same drawing so for angle theta we have the angle 30 degrees we have that the weight of the block is w the weight of the strut is also w and the tension on the string is t so what i'm going to do is is to draw a force diagram on the struid.

02:48 So here, at this point, we have the weight force coming from the block.

02:58 Weight force w.

03:01 At this point, we also have the tension perpendicular to the weight force.

03:07 We have the weight force acting on the strut itself.

03:11 It acts on the center of mass of the strut.

03:13 And finally we have the force of the pivot exerts on the strut.

03:19 It points somewhere in this direction.

03:23 Okay, i'm just drawing a general direction to the northeast and also i'm going to write that w, i'm going to set up a coordinate system such that the y axis points to the north, the x axis points to the right, so y, i'm sorry, w is equal to minus w, j, t is equal to minus t i, where w and j are the magnitude of the forces.

04:05 So all we need to do is to find both the tension, magnitude of the tension, and also the direction and magnitude of the force f.

04:16 And here in my drawing, it seems like the force necessarily is pointing in the direction of the direction of.

04:20 Of the strut.

04:21 Since that's not true, let me just be a little clearer, let's say that points in a general direction.

04:27 So we have two equations.

04:29 The first one that i'm going to apply is the equation for the torque.

04:35 So the sum of the torques is zero.

04:37 Notice that the strut, if it were to rotate, it will rotate around the pivot.

04:42 So the force of the pivot exerts on the strut will not be counted in the torque equation because the distance between the point where the force is applied and the central rotation is zero.

04:57 So in this angle here is 30 degrees.

05:02 This means that this one is 60.

05:05 And by the right -hand rule, we can see that the direction of the torque coming from the tension on the rope is equal.

05:14 Yeah, the direction is in the z direction, even though the z -direction, is coming out of the screen.

05:25 And i'm going to consider this direction, the positive direction.

05:28 The negative z direction, i'm going to consider the negative direction.

05:30 So the torque coming from t is equal to t times the length of the rope.

05:41 I'm going to call it l, tl, tl times the sign of 60.

05:52 The torque that is exerted by the 1 weight force coming from the block, i'm going to call it tau w1, is equal to omega l times sine of 60.

06:09 Actually, up here i wrote that the sine of 60 should have been sign of 30.

06:13 And there's a minus sign here because by the right hand rule you can see that the torque coming from the weight force is pointing in the negative z direction...

Please give Ace some feedback