Question

A uniform, equilateral triangle of side length \(2.00 \mathrm{~m}\) and weight \(4.00 \cdot 10^{3} \mathrm{~N}\) is placed across a gap. One point is on the north end of the gap, and the opposite side is on the south end. Find the force on each side.

Short answer

Answer: The force acting on the north side of the gap is 2.67 * 10^3 N, and the force acting on the south side is 1.33 * 10^3 N.

Step-by-step solution

Identifying Forces

We will begin by identifying the forces acting on the triangle: 1) The weight (W) of the triangle acting downward at the centroid. 2) North support force (N) acting vertically upward at the north end of the gap. 3) South support force (S) acting vertically upward at the intersection of the south end of the gap and the base of the triangle. We will set up the equilibrium equations for the sum of vertical forces and the sum of moments around the north point (i.e., N).

Sum of Vertical Forces

The sum of the vertical forces equals zero: N + S - W = 0 Where N is the north support force, S is the south support force, and W is the weight of the triangle. We will use this equation to find a relationship between N and S.

Sum of Moments

The sum of moments around the north point is zero. Taking the north support point as the axis of rotation, we have: Moment_due_to_S + Moment_due_to_W = 0 We need to find the perpendicular distance from the south support force to the north support point and the distance from the centroid to the north support point. Using geometry, we can find these distances: Distance_from_S_to_N = 2.00 m (side of the equilateral triangle) Distance_from_centroid_to_N = (2.00 m) * (1/3) Moment_due_to_S = S * Distance_from_S_to_N Moment_due_to_W = W * Distance_from_centroid_to_N Substitute the values and solve for S: S * (2.00 m) = W * (2.00 m) * (1/3)

Solve for North and South Forces

Now, using the equations from step 2 and step 3, we can find the north and south forces: N + S - W = 0 S * (2.00 m) = W * (2.00 m) * (1/3) First, we can find the south force S: S = W * (1/3) S = (4.00 * 10^3 N) * (1/3) S = 1.33 * 10^3 N Then, using the equation from step 2, we can find the north force N: N = W - S N = (4.00 * 10^3 N) - (1.33 * 10^3 N) N = 2.67 * 10^3 N Thus, the north support force is 2.67 * 10^3 N, and the south support force is 1.33 * 10^3 N.

Key concepts

Static Equilibrium

When an object is at rest or moving at a constant velocity, it is said to be in static equilibrium. This means that all the forces and torques (moments) acting upon it are balanced. In our exercise, the equilateral triangle, despite being subject to gravitational pull, doesn't move or rotate. This is because the support forces at the north and south ends provide an equal upward force that counteracts the triangle's weight, maintaining its state of equilibrium.

For students, understanding static equilibrium involves recognizing that for an object to be in this state, two main conditions must be satisfied: The sum of all horizontal forces and the sum of all vertical forces acting on the object must be zero, and the sum of all moments about any axis must also be zero.

Torque and Moments

Torque, or moment of force, is the rotational equivalent of force. It is a measure of how much a force causes an object to rotate around an axis. The amount of torque can be calculated as the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force, represented by the formula torque (τ) = force (F) × distance (d).

In the given exercise, torque is crucial in understanding how the forces at the supports influence the rotation of the triangle. We calculate moments around the north support point to ensure that the triangle doesn't tip over, therefore satisfying the static equilibrium condition for moments. This is a pivotal concept in physics and engineering, often requiring careful calculation to ensure the stability of structures.

Center of Mass

The center of mass of an object is the point at which its entire mass can be considered to be concentrated. For uniform symmetrical objects, like the equilateral triangle in the problem, the center of mass is at the geometric center. In our case, the center of mass is important since it is the point where the weight of the triangle acts downward.

The position of the center of mass affects where and how forces should be applied to achieve equilibrium. For students learning about equilibrium, knowing how to locate the center of mass helps in solving problems involving balancing and stability of objects.

Support Forces

Support forces are the forces exerted by a surface or point that upholds the weight of an object and prevents it from failing due to gravity. In our exercise, there are two such forces: one at the north end and one at the south end of the triangle. These forces react to the triangle's weight to maintain it in static equilibrium.

Understanding the role of support forces involves determining the magnitude and direction of these forces. It requires a clear insight into how they contribute to the overall balance of the system. This is a foundational concept for anyone dealing with physical structures or mechanical systems.

Equilateral Triangle Physics

The properties of an equilateral triangle play an essential role in physics problems that involve this shape. An equilateral triangle is a triangle in which all three sides are of equal length, and all angles are equally 60 degrees. The symmetry of this shape simplifies calculations and assumptions.

For the given problem, knowing that the triangle is equilateral allows us to easily determine distances from the supports to the centroid, which is crucial in calculating torques and moments. Recognizing the inherent symmetry in an equilateral triangle aids in understanding the distribution of forces and can be used to solve more complex problems involving this shape.