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When the moment of force is maximum, then what is the angle between force and position vector of the force? (A) \(90^{\circ}\) (B) \(0^{\circ}\) (C) \(30^{\circ}\) (D) \(45^{\circ}\)

Short Answer

Expert verified
(A) \(90^{\circ}\)

Step by step solution

01

Moment of Force Formula

The moment of force, also known as torque, can be calculated using the following formula: \( { \tau = \vec{r} \times \vec{F}} \) where \(\tau\) is the torque, \(\vec{r}\) is the position vector, and \(\vec{F}\) is the force vector. The magnitude of the torque can be expressed as: \( {|\tau| = rF \sin \theta} \) where \(||\) indicates absolute value, \(r\) is the magnitude of position vector, \(F\) is the magnitude of the force vector, and \(\theta\) is the angle between \(\vec{r}\) and \(\vec{F}\).
02

Finding the Maximum Torque

We want to find the angle at which the torque is maximum. To do this, we will look at how the angle (\(\theta\)) influences the torque expression: \(rF \sin \theta\). Notice that the sine function has a maximum value of 1 when the angle is \({90}^{\circ}\). In other words, the maximum possible value of the sine function is 1, which occurs when the angle is \({90}^{\circ}\).
03

Conclusion

Putting it all together, when \(\theta = {90}^{\circ}\), the maximum torque is obtained for any given position vector and force vector. Therefore, the answer is: (A) \(90^{\circ}\)

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Most popular questions from this chapter

In the experiment of balancing moments, suppose the fulcrum is at the $60 \mathrm{~cm}\( mark, and a known mass of \)2 \mathrm{~kg}$ is used on the longer arm. The greatest mass of \(\mathrm{m}\) which can be balanced against $2 \mathrm{~kg}$ such that the minimum distance of either of the masses from the fulcrum is at least \(10 \mathrm{~cm}\). (Neglect mass of meter scale.) What will be the value of \(\mathrm{m}\) ? (A) \(12 \mathrm{~kg}\) (B) \(4 \mathrm{~kg}\) (C) \(6 \mathrm{~kg}\) (D) \(8 \mathrm{~kg}\)

A force \(2 \mathrm{i} \wedge+3 \mathrm{j} \wedge\) acts about an axis at a position vector \((\mathrm{j} \wedge+\mathrm{k} \wedge)\) from the axis, then what is the torque due to the force about the axis? (A) \(\sqrt{19} \mathrm{Nm}\) (B) \(\sqrt{13} \mathrm{Nm}\) (C) \(\sqrt{15 \mathrm{Nm}}\) (D) \(\sqrt{17} \mathrm{Nm}\)

When a meter scale is balanced above a wedge, \(1 \mathrm{~kg}\) mass is hung at \(10 \mathrm{~cm}\) mark and a \(2 \mathrm{~kg}\) mass is hang at the $85 \mathrm{~cm}$ mark. To which mark on the meter scale, the fulcrum be shifted (Neglect mass of meter scale) to balance the scale? (A) \(70 \mathrm{~cm}\) (B) \(50 \mathrm{~cm}\) (C) \(60 \mathrm{~cm}\) (D) \(65 \mathrm{~cm}\)

When the "wedge and scale" experiment is performed at the equator, we get \(m=M\left(y_{e} / x_{e}\right)\). If the same experiment is performed at the poles, then what is the right equation ? (A) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right) \cdot\left(\mathrm{R}_{\mathrm{e}} / \mathrm{R}_{\mathrm{p}}\right)$ (B) \(m=M\left(y_{e} / x_{e}\right) \cdot\left(R_{p} / R_{e}\right)\) (C) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right) \cdot\left(\mathrm{R}_{\mathrm{p}} / \mathrm{R}_{\mathrm{e}}\right)^{2}$ (D) $\mathrm{m}=\mathrm{M}\left(\mathrm{y}_{\mathrm{e}} / \mathrm{x}_{\mathrm{e}}\right)$ Where \(\mathrm{R}_{\mathrm{e}}\) and \(\mathrm{R}_{\mathrm{p}}\) are the equatorial and polar radius of the earth.

The wedge is kept below the \(60 \mathrm{~cm}\) mark on the meter scale. Known masses of \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are hung at the $20 \mathrm{~cm}\( and \)30 \mathrm{~cm}\( mark respectively. Where will a \)4 \mathrm{~kg}$ mass be hung on the meter scale to balance it ? (Neglect mass of meter scale.) (A) \(85 \mathrm{~cm}\) (B) \(90 \mathrm{~cm}\) (C) \(70 \mathrm{~cm}\) (D) \(75 \mathrm{~cm}\)

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