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Chapter 9: Problem 9

What is the torque about the origin on a particle located at \(x=1.5 \mathrm{~m}, \quad y=-2.0 \mathrm{~m}, z=1.6 \mathrm{~m}\) due to a force \(\overrightarrow{\mathbf{F}}=\) \((3.5 \mathrm{~N}) \hat{\mathbf{i}}-(2.4 \mathrm{~N}) \hat{\mathbf{j}}+(4.3 \mathrm{~N}) \hat{\mathbf{k}} ?\) Express your result in unit vector notation.

Short Answer

Expert verified
The torque of the force on the particle about the origin, expressed in unit vector notation, is \( \tau = \overrightarrow{r}\times\overrightarrow{F} = -8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm \).

Step by step solution

01

Identify the Position and Force Vectors

First, identify the position and force vectors from the exercise. The position vector \(\overrightarrow{r}\) from the origin to the particle in unit vector notation is \(1.5\hat{i} - 2.0\hat{j} + 1.6\hat{k}\) and the force vector \(\overrightarrow{F}\) is \((3.5 N)\hat{i} - (2.4 N)\hat{j} + (4.3 N)\hat{k}\).
02

Compute the cross product

Calculate the cross product of the position vector and the force vector, that is \(\overrightarrow{r}\times\overrightarrow{F}\). This is done by determining the determinant of the following matrix:\[\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \1.5 & -2.0 & 1.6 \3.5 & -2.4 & 4.3\end{vmatrix}.\] Expanding the determinant along the first row, we get:\[\overrightarrow{r}\times\overrightarrow{F} = (-2.0*4.3+1.6*-2.4)i - (1.5*4.3-1.6*3.5)j + (1.5*-2.4+3.5*-2.0)k.\] Calculating gives: \(-8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm\).
03

Express Result in Unit Vector Notation

The final step is to express the result in unit vector notation. So, the torque of the force on the particle is \( \tau = \overrightarrow{r}\times\overrightarrow{F} = -8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm \).

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Most popular questions from this chapter

A pulley having a rotational inertia of \(1.14 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(9.88 \mathrm{~cm}\) is acted on by a force, applied tangentially at its rim, that varies in time as \(F=A t+B t^{2}\), where \(A=0.496 \mathrm{~N} / \mathrm{s}\) and \(B=0.305 \mathrm{~N} / \mathrm{s}^{2} .\) If the pulley was initially at rest, find its angular speed after \(3.60 \mathrm{~s}\).

Vector \(\overrightarrow{\mathbf{a}}\) has magnitude \(3.20\) units and lies in the \(y z\) plane \(63.0^{\circ}\) from the \(+y\) axis with a positive \(z\) component. Vector \(\overrightarrow{\mathbf{b}}\) has magnitude \(1.40\) units and lies in the \(x z\) plane \(48.0^{\circ}\) from the \(+x\) axis with a positive \(z\) component. Find \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\).

Figure \(9-54\) shows the massive shield door at a neutron test facility at Lawrence Livermore Laboratory; this is the world's heaviest hinged door. The door has a mass of \(44,000 \mathrm{~kg}\), a rotational inertia about its hinge line of \(8.7 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and a width of \(2.4 \mathrm{~m}\). What steady force, applied at its outer edge at right angles to the door, can move it from rest through an angle of \(90^{\circ}\) in \(30 \mathrm{~s}\) ?

A wheel in the form of a uniform disk of radius \(23.0 \mathrm{~cm}\) and mass \(1.40 \mathrm{~kg}\) is turning at 840 rev/min in frictionless bearings. To stop the wheel, a brake pad is pressed against the rim of the wheel with a radially directed force of \(130 \mathrm{~N}\). The wheel makes \(2.80\) revolutions in coming to a stop. Find the coefficient of friction between the brake pad and the rim of the wheel.

In an Atwood's machine one block has a mass of \(512 \mathrm{~g}\) and the other a mass of \(463 \mathrm{~g}\). The pulley, which is mounted in horizontal frictionless bearings, has a radius of \(4.90 \mathrm{~cm}\). When released from rest, the heavier block is observed to fall \(76.5 \mathrm{~cm}\) in \(5.11\) s. Calculate the rotational inertia of the pulley.
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