A uniform metal beam of length L weighs 200 N and holds a...

A uniform metal beam of length $L$ weighs $200 \mathrm{~N}$ and holds a $450-\mathrm{N}$ object as shown in Fig. $5-3$. Find the magnitudes of the forces exerted on the beam by the two supports at its ends. Assume the lengths are exact. Rather than draw a separate free-body diagram, we show the forces on the object being considered (the beam) in Fig. $5-3$. Because the beam is uniform, its center of gravity is at its geometric center. Thus, the weight of the beam ( $200 \mathrm{~N}$ ) is shown acting downward at the beam's center. The forces $F_{1}$ and $F_{2}$ are exerted on the beam by the supports. Because there are no $x$ -directed forces acting on the beam, we have only two equations to write for this equilibrium situation: $\sum F_{y}=0$ and $\sum \tau=0$. $$ +\uparrow \sum F_{y}=0 \quad \text { becomes } \quad F_{1}+F_{2}-200 \mathrm{~N}-450 \mathrm{~N}=0 $$ Before the torque equation is written, an axis must be chosen. We choose it at $A$, so that the unknown force $F_{1}$ will pass through it and exert no torque. The torque equation is then $$ \text { \&+ } \sum \tau_{A}=-(L / 2)(200 \mathrm{~N})\left(\sin 90^{\circ}\right)-(3 L / 4)(450 \mathrm{~N})\left(\sin 90^{\circ}\right)+L F_{2} \sin 90^{\circ}=0 $$ Dividing through the equation by $L$ and solving for $F_{2}$, we find that $F_{2}=438 \mathrm{~N}$. To determine $F_{1}$, substitute the value of $F_{2}$ in the force equation, thereby obtaining $F_{1}=212 \mathrm{~N}$.

A uniform metal beam of length $L$ weighs $200 \mathrm{~N}$ and holds a $450-\mathrm{N}$ object as shown in Fig. $5-3$. Find the magnitudes of the forces exerted on the beam by the two supports at its ends. Assume the lengths are exact.

Rather than draw a separate free-body diagram, we show the forces on the object being considered (the beam) in Fig. $5-3$. Because the beam is uniform, its center of gravity is at its geometric center. Thus, the weight of the beam ( $200 \mathrm{~N}$ ) is shown acting downward at the beam's center. The forces $F_{1}$ and $F_{2}$ are exerted on the beam by the supports. Because there are no $x$ -directed forces acting on the beam, we have only two equations to write for this equilibrium situation: $\sum F_{y}=0$ and $\sum \tau=0$.
$$
+\uparrow \sum F_{y}=0 \quad \text { becomes } \quad F_{1}+F_{2}-200 \mathrm{~N}-450 \mathrm{~N}=0
$$
Before the torque equation is written, an axis must be chosen. We choose it at $A$, so that the unknown force $F_{1}$ will pass through it and exert no torque. The torque equation is then
$$
\text { \&+ } \sum \tau_{A}=-(L / 2)(200 \mathrm{~N})\left(\sin 90^{\circ}\right)-(3 L / 4)(450 \mathrm{~N})\left(\sin 90^{\circ}\right)+L F_{2} \sin 90^{\circ}=0
$$
Dividing through the equation by $L$ and solving for $F_{2}$, we find that $F_{2}=438 \mathrm{~N}$.
To determine $F_{1}$, substitute the value of $F_{2}$ in the force equation, thereby obtaining $F_{1}=212 \mathrm{~N}$.

Key Concepts

Static Equilibrium

Static equilibrium refers to the condition in which an object remains at rest or moves with constant velocity due to the net force and net moment (torque) on it being zero. This concept is fundamental in mechanics as it sets the stage for analyzing forces acting on a body by ensuring that the summation of all forces and the summation of all torques about any point are both equal to zero.

Free-Body Diagram (FBD)

A free-body diagram is a simplified representation of a system or component that isolates the object of interest and shows all the external forces acting on it. This technique is crucial because it allows one to visualize and balance all forces and moments, making it easier to set up the equilibrium equations needed for solving statics problems.

Torque (Moment) and Lever Arm

Torque, also known as moment, is the measure of the tendency of a force to rotate an object about an axis. It is calculated by multiplying the force by the perpendicular distance (lever arm) from the axis of rotation to the line of action of the force. Understanding torque is essential for setting up rotational equilibrium conditions, especially when forces produce turning effects about a pivot.

Center of Mass in Uniform Bodies

For uniform bodies, the center of mass is located at the geometric center, meaning that the weight of the body is evenly distributed with its gravitational force acting as if it were concentrated at that point. Recognizing this concept simplifies analysis by allowing one to replace a distributed weight with a single force acting at the center of mass.

Choice of Pivot or Axis in Equilibrium Analysis

The selection of a pivot point or axis for setting up the torque equilibrium equation is a strategic decision in statics problems. By choosing a point where one or more unknown forces act, those forces exert no moment about that point, thereby simplifying the torque equation and reducing the number of unknowns. This method is widely used to systematically solve for unknown forces in equilibrium situations.

Transcript

00:01 Hello everyone, let's solve the following question.

00:02 In this question, we are given a metal beam which weighs 200 newton.

00:06 At this point, there is an object whose rate is 4 .50 newton and it's hanging at this point.

00:12 Now, the whole length of this beam is l.

00:14 They have divided the whole length into three sections.

00:17 This length is l by 2 from a to center and this length is l by 4 and this length is also l by 4.

00:24 Now we are taking this point as the center of the beam, this point as the center of beam because this beam is uniform.

00:32 This beam is uniform.

00:35 So the whole weight will act as a geometrical center of the beam.

00:39 What we have to find in this question, we have to find the force f1 and force f2 which will be applied on the ends of the beam by these two sports.

00:48 So for finding the f1, f2, we can see in this figure that there are no forces which are.

00:54 Acting in the x -axis along the x -axis there are no forces forces are zero so if we take the sum of forces in the y direction it will be equals to zero because this beam is in the equilibrium position further if we take a as the fixed point this a as the fixed point that we can also find torque about this point so torque smission of torque about a point should be equals to 0.

01:21 So first of all, we will expand the first equation.

01:24 Smission of f by will be equals to 0.

01:27 Smission of f by will be equals to 0.

01:31 So what will be smission of f by? f by forces acting along the y direction.

01:36 F1 and f2, both are in the same direction.

01:39 So we will write f1 plus f2...

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