- 1(current)
Q. No.The value of \(M\), as shown, for which the rod will be in equilibrium is:
| 1. | \(1\) kg | 2. | \(2\) kg |
| 3. | \(4\) kg | 4. | \(6\) kg |
In the figure given below, \(O\) is the centre of an equilateral triangle \(ABC\) and \(\vec{F_{1}} ,\vec F_{2}, \vec F_{3}\) are three forces acting along the sides \(AB\), \(BC\) and \(AC\). What should be the magnitude of \(\vec{F_{3}}\) so that total torque about \(O\) is zero?
1. \(\left|\vec{F_{3}}\right|= \left|\vec{F_{1}}\right|+\left|\vec{F_{2}}\right|\)
2. \(\left|\vec{F_{3}}\right|= \left|\vec{F_{1}}\right|-\left|\vec{F_{2}}\right|\)
3. \(\left|\vec{F_{3}}\right|= \vec{F_{1}}+2\vec{F_{2}}\)
4. Not possible
A wheel with a radius of \(20\) cm has forces applied to it as shown in the figure. The torque produced by the forces of \(4\) N at \(A\), \(8~\)N at \(B\), \(6\) N at \(C\), and \(9~\)N at \(D\), at the angles indicated, is:
1. \(5.4\) N-m anticlockwise
2. \(1.80\) N-m clockwise
3. \(2.0\) N-m clockwise
4. \(3.6\) N-m clockwise
| 1. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
| 2. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}) \) N-m |
| 3. | \(\vec{\tau}=(17 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
| 4. | \(\vec{\tau}=(-41 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+16 \hat{\mathrm{k}})\) N-m |
| 1. | \(wx \over d\) | 2. | \(wd \over x\) |
| 3. | \(w(d-x) \over x\) | 4. | \(w(d-x) \over d\) |
For L = 3.0 m, the total torque about pivot A provided by the forces as shown in the figure is:
| 1. | 210 Nm | 2. | 140 Nm |
| 3. | 95 Nm | 4. | 75 Nm |
- 1(current)
Q. No.
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