particle of mass $$\(2.5 \mathrm{~kg}\)$$ is held in equilibrium on a rough plane inclined at $$\(20^{\circ}\)$$ to the horizontal by a force of magnitude $$\(T \mathrm{~N}\)$$ making an angle of $$\(60^{\circ}\)$$ with a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is $$\(0.3\)$$. Find the greatest and least possible values of $$\(T\)$$. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Exam No：9709_s20_qp_42 Year：2020 Question No：3
Answer:
\(T \sin 60+R=25 \cos 20\)
Attempt at resolving in any direction
\(T \cos 60=F+25 \sin 20\)
\(T \cos 60+F=25 \sin 20\)
Use of \(F=\mu R\)
\(T \cos 60=25 \sin 20 \pm 0.3(25 \cos 20-T \sin 60)\)
\(T=\frac{25 \sin 20 \pm 0.3 \times 25 \cos 20}{\cos 60 \pm 0.3 \sin 60}\)
\(T=6.26\)
\(T=20.5\)
Attempt at resolving in any direction
\(T \cos 60=F+25 \sin 20\)
\(T \cos 60+F=25 \sin 20\)
Use of \(F=\mu R\)
\(T \cos 60=25 \sin 20 \pm 0.3(25 \cos 20-T \sin 60)\)
\(T=\frac{25 \sin 20 \pm 0.3 \times 25 \cos 20}{\cos 60 \pm 0.3 \sin 60}\)
\(T=6.26\)
\(T=20.5\)
Knowledge points:
4.1.1 identify the forces acting in a given situation； e.g. by drawing a force diagram.
4.1.2 understand the vector nature of force, and find and use components and resultants； Calculations are always required, not approximate solutions by scale drawing.
4.1.3 use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero （Solutions by resolving are usually expected, but equivalent methods (e.g. triangle of forces, Lami’s Theorem, where suitable) are also acceptable; these other methods are not required knowledge, and will not be referred to in questions.）
4.1.4 understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component
4.1.6 understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = nR or F G nR, as appropriate
Solution:
Download APP for more features
1. Tons of answers.
2. Smarter Al tools enhance your learning journey.
IOS
Download
Download
Android
Download
Download
Google Play
Download
Download