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Chapter 4: Problem 48

A rectangular block of width \(w=116.5 \mathrm{~cm},\) depth \(d=164.8 \mathrm{~cm}\) and height \(h=105.1 \mathrm{~cm}\) is cut diagonally from one upper corner to the opposing lower corners so that a triangular surface is generated, as shown in the figure. A paperweight of mass \(m=16.93 \mathrm{~kg}\) is sliding down the incline without friction. What is the magnitude of the acceleration that the paperweight experiences?

Short Answer

Expert verified
Answer: The magnitude of the acceleration that the paperweight experiences while sliding down the incline without friction is approximately \(4.50\mathrm{~m/s^2}\).

Step by step solution

01

Find the length of the diagonal

To find the length of the diagonal, we will use the Pythagorean theorem in 3D space which can be stated as: \(l^2 = w^2 + d^2 + h^2\). Given the dimensions of the rectangular block, we can plug the values and find the diagonal length. \(l^2 = 116.5^2 + 164.8^2 + 105.1^2\) Calculating the diagonal length: \(l = \sqrt{116.5^2 + 164.8^2 + 105.1^2} \approx 230.37 \mathrm{~cm}\)
02

Find the angle of the incline

Next, we need to find the angle at which the incline is tilted. We can use the height \(h\) and the diagonal length \(l\) to find the angle using trigonometry. Let \(\theta\) be the angle between the diagonal \(l\) and the base of the rectangular block. Using the sine function, we can write: \(\sin{\theta} = \frac{h}{l}\) Now, we can find the angle \(\theta\): \(\theta = \arcsin{\frac{105.1}{230.37}} \approx 27.51^\circ\)
03

Find the acceleration of the paperweight along the incline

Finally, we can find the acceleration of the paperweight sliding down the incline using the gravitational force and the angle of the incline. We know that the force acting along the incline is: \(F = mg\sin{\theta}\) Now, we can use Newton's second law \(F = ma\) to find the acceleration: \(16.93\mathrm{~kg} \cdot a = 16.93\mathrm{~kg} \cdot 9.81\mathrm{~m/s^2} \cdot \sin{27.51^\circ}\) Divide both sides by the mass \(m\): \(a = 9.81\mathrm{~m/s^2} \cdot \sin{27.51^\circ} \approx 4.50\mathrm{~m/s^2}\) The magnitude of the acceleration that the paperweight experiences while sliding down the incline without friction is approximately \(4.50\mathrm{~m/s^2}\).

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Most popular questions from this chapter

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