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Chapter 11: Problem 32

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

Expert verified
Answer: The force on the north end (vertex): N 鈮� (16鈭�3 / 3) x 10^3 N and the force on the south end (center of the side): S 鈮� 4.00 x 10^3 N - (16鈭�3 / 3) x 10^3 N.

Step by step solution

01

Identify Forces Acting on the Triangle

To find the forces acting on the triangle, we first need to identify all the external forces acting on it. There are three forces acting on the triangle: The weight W (4.00 x 10^3 N) acting downwards at the centroid of the triangle, the force N acting upwards at the north end, and the force S acting upwards at the south end.
02

Calculate the Centroid of the Equilateral Triangle

In an equilateral triangle, the centroid is the same as the center of the triangle, which can be found at the intersection of the medians. The length of each median in an equilateral triangle is \[\frac{2\sqrt{3}}{3}\] times the side length, so for our triangle with side length of 2.00 m, the centroid is located at a height of \[h_c = \frac{2\sqrt{3}}{3} \times 2.00 \mathrm{~m} = \frac{4\sqrt{3}}{3} \mathrm{~m}\] below the vertex on the north end.
03

Equilibrium of Forces

Since the triangle is in equilibrium, the total vertical force acting on it must be zero. Thus, we can write the equilibrium of forces equation as follows: \[N + S = W\]
04

Calculate the Distance between the Forces and the Center of Mass

Let's find the horizontal distance from the centroid of the triangle to the north end (vertex) and the south end (center of the side). We can use trigonometry to determine these distances. From the north end, the distance to the centroid is d1 = (2.00 m / 2) = 1.00 m. From the south end, the distance to the centroid is d2 = 0 (since the force is applied at the center of the side).
05

Equilibrium of Moments

Considering the equilibrium of moments about the centroid, we have the following equation: \[(N)d_{1} = (W)h_{c}\] Solving for the force N, we get: \[N = \frac{(W)h_{c}}{d_{1}}\]
06

Calculate the Force N

Now, we can substitute the known values for W, hc, and d1 to find the force N: \[N = \frac{(4.00\times10^{3}\mathrm{~N})(\frac{4\sqrt{3}}{3}\mathrm{~m})}{(1.00\mathrm{~m})}\] \[N = \frac{16\sqrt{3}}{3} \times 10^{3} \mathrm{~N}\]
07

Calculate the Force S

Now, we can use the equilibrium of forces equation N + S = W to find the force S: \[S = W - N\] Substituting the known values for W and N, we get: \[S = 4.00 \times 10^{3}\mathrm{~N} - \frac{16\sqrt{3}}{3} \times 10^{3}\mathrm{~N}\]
08

Present the Final Results

The force on each side of the triangle is as follows: Force on the north end (vertex): N = (16鈭�3 / 3) x 10^3 N Force on the south end (center of the side): S = 4.00 x 10^3 N - (16鈭�3 / 3) x 10^3 N

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Most popular questions from this chapter

Why didn't the ancient Egyptians build their pyramids upside-down? In other words, use force and center-of-mass principles to explain why it is more advantageous to construct buildings with broad bases and narrow tops than the other way around.

A 600.0-N bricklayer is \(1.5 \mathrm{~m}\) from one end of a uniform scaffold that is \(7.0 \mathrm{~m}\) long and weighs \(800.0 \mathrm{~N}\). A pile of bricks weighing \(500.0 \mathrm{~N}\) is \(3.0 \mathrm{~m}\) from the same end of the scaffold. If the scaffold is supported at both ends, calculate the force on each end.

A boy weighing \(60.0 \mathrm{lb}\) is playing on a plank. The plank weighs \(30.0 \mathrm{lb}\), is uniform, is \(8.00 \mathrm{ft}\) long, and lies on two supports, one \(2.00 \mathrm{ft}\) from the left end and the other \(2.00 \mathrm{ft}\) from the right end. a) If the boy is \(3.00 \mathrm{ft}\) from the left end, what force is exerted by each support? b) The boy moves toward the right end. How far can he go before the plank will tip?

A 2.00 -m-long diving board of mass \(12.0 \mathrm{~kg}\) is \(3.00 \mathrm{~m}\) above the water. It has two attachments holding it in place. One is located at the very back end of the board, and the other is \(25.0 \mathrm{~cm}\) away from that end. a) Assuming that the board has uniform density, find the forces acting on each attachment (take the downward direction to be positive). b) If a diver of mass \(65.0 \mathrm{~kg}\) is standing on the front end, what are the forces acting on the two attachments?

A construction supervisor of mass \(M=92.1 \mathrm{~kg}\) is standing on a board of mass \(m=27.5 \mathrm{~kg} .\) Two sawhorses at a distance \(\ell=3.70 \mathrm{~m}\) apart support the board. If the man stands a distance \(x_{1}=1.07 \mathrm{~m}\) away from the left-hand sawhorse as shown in the figure, what is the force that the board exerts on that sawhorse?
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