Application to Mechanics
A particle is acted upon by forces of \((3i - 2j + k)\) N, \((7i + 4j + 3k)\) N and \((-5i - 3j)\) N.
a) Work out the resultant force R.
b) Find the exact magnitude of the resultant force
a)
- \(3i+7i-5i=5i\)
- \(-2j+4j-3j=-j\)
- \(k+3k=4k\)
- Resultant force: \(\begin{pmatrix}5i \\ -j \\ 4k\end{pmatrix}\)
b)
- \(\sqrt{ 5^2+1^2+4^2 }=\sqrt{ 25+1+16 }=\sqrt{ 42 }\)
A particle, initially at rest, is acted upon by a force that causes the particle to accelerate at \((4i - 2j + 3k)\text{ }ms^{-2}\) for 2 seconds. Work out the distance travelled by the particle.
- \(a=\sqrt{ 4^2+2^2+3^2 }=\sqrt{ 29 }\)
- \(s=ut+\frac{1}{2}at^2\)
- \(s=(0\times4)+\left( \frac{1}{2}\times \sqrt{ 29 }\times4=2\sqrt{ 29 } \right)\)
- \(s=2\sqrt{ 29 }\)
A body of mass 4 kg is moving with a constant velocity when it is acted upon by a force of \((2i - 5j + 3k)\) N.
a) Find the acceleration of the body while the force acts.
b) Find the magnitude of this acceleration to 3 s.f.
a)
- \(F=ma\)
- \(a=\frac{F}{m}\)
- \(a=\frac{2}{4}i-\frac{5}{4}j+\frac{3}{4}k\)
- \(a=\frac{1}{2}i-\frac{5}{4}j+\frac{3}{4}k\)
b)
- \(F=ma\)
- \(a=\frac{F}{m}\)
- \(a=\frac{\sqrt{ 2^2+5^2+3^2 }}{4}=\frac{\sqrt{ 38 }}{4}\)
A particle of mass \(6kg\) is acted on by two forces, \(F_{1}\) and \(F_{2}\). Given that F1 = \((7i + 3j + k)\) N, and that the particle is accelerating at \((2i - k)\text{ }ms^{-2}\), find \(F_{2}\), giving your answer in the form \((pi + qj + rk)\) N.
- \(F_{1}+F_{2}=ma\)
- \((7i+3j+k)+F_{2}=12i-6k\)
- \(F_{2}=(5i-3j-7k)N\)
A particle of mass 2 kg is in static equilibrium and is acted upon by three forces: \(F_{1} = (i - j - 2k)\) N, \(F_{2} = (-i + 3j + bk)\) N and \(F_{3} = (aj - 2k)\) N
- a) Find the values of the constants \(a\) and \(b\).
- \(F_{2}\) is removed. Work out:
- b) the value of the resultant force \(R\)
- c) the acceleration of the particle, giving your answer in the form \((pi + qj + rk)\text{ }ms^{-2}\)
- d) the magnitude of this acceleration
- e) the angle the acceleration vector makes with the unit vector \(\mathbf{j}\).
a)
- To balance so all values are 0:
- \(a=-2\)
- \(b=4\)
b)
- \(R=(i-3j-4k)\)