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A body of mass 10 kg is held in place on a smooth plane inclined at 30° to the horizontal by a tension force, \(T\) newtons, acting parallel to the plane.
Assuming that the acceleration due to gravity is 9.8 m s\(^{-2}\), find the value of \(T\) in newtons. (2 marks)
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A 6-kilogram mass is placed on a frictionless plane inclined 30° to the horizontal.
The mass is connected to another 10-kilogram mass by a light, inextensible string via a pulley, as shown in the diagram
The 6-kilogram mass is released from rest and slides up the plane.
Given the acceleration due to gravity is \(g\) m s\(^{-2}\), express the velocity, \(v\), of the 6 kilogram mass in terms of \(t\) as it moves up the plane. (3 marks)
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\(\text{Resolving forces on the 6 kg mass:}\)
\(\text{Let} \ \ F_d=\text{force down slope:}\)
| \(\sin 30^{\circ}\) | \(=\dfrac{F_d}{6 g}\) |
| \(F_d\) | \(=6 g\, \sin 30=3 g\) |
\(\text{Using} \ \ \sum F=m a\):
\(T-3 g=6 a\ \ldots\ (1)\)
\(\text{Consider 10 kg mass:}\)
\(10 g-T=10 a\ \ldots\ (2)\)
\(\text{Add (1) + (2):}\)
\(7 g=16 a \ \Rightarrow \ a=\dfrac{7}{16} g\)
\(v=\displaystyle \int \dfrac{7}{16}g \, dt=\dfrac{7}{16} g t+c\)
\(\text{When} \ \ t=0 \quad v=0 \ \Rightarrow \ c=0\)
\(\therefore v=\dfrac{7}{16}gt\)
An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is \(g \ \text{ms} ^{-2}\) and the velocity of the object down the slope is \(v \ \text{ms} ^{-1}\).
As well as the force due to gravity, the object is acted on by two forces, one of magnitude \(2 v\) newtons and one of magnitude \(2 v^2\) newtons, both acting up the slope.
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a.
| \(\Sigma F\ \text{(down slope) }\) | \(=5 g\ \cos 30^{\circ}-2 v-2 v^2\) |
| \(=\dfrac{5 \sqrt{3}}{2} g-2 v-2 v^2 \ \ \text{newtons }\) |
b. \(\text{Constant speed} \ \Rightarrow \ \ \Sigma F=0\)
| \(\dfrac{5 \sqrt{3}}{2} \times 10\) | \(=2 v+2 v^2\) | |
| \(0\) | \(=2 v^2+2 v-25 \sqrt{3}\) |
\(v=\dfrac{-2 \pm \sqrt{4+4(2)(25 \sqrt{3})}}{4} =4.1798 \ldots \text { or }-5.1798 \ldots\)
\(\text{Since object is moving down the slope,}\ \ v \gt 0.\)
\(\therefore v=4.1798 = 4.2 \ \text{ms}^{-1} \ \ \text{(1 d.p.)}\)
A 10 kg mass is placed on a smooth plane that is inclined at 30° to the horizontal, as shown in the diagram below. A force, `F` is applied to the mass up the slope and parallel to the slope so that when released, the mass remains at rest.
If the acceleration due to gravity, `g`, is `9.8\ text(m s)^{-2}`, determine the magnitude of the force, in newtons, acting up the slope. (2 marks)
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| `sin 30^@` | `= F/(10g)` |
| `F` | `=10g\ sin30^@` |
| `F` | `= 10 xx 9.8 xx 1/2 = 49\ text{N}` |
An object of mass `m` kilograms slides down a smooth slope that is inclined at an angle of `theta^@` to the horizontal, where `0^@ < theta^@ < 45^@`. The acceleration of the object down the slope is `a\ text(ms)^(-2), a > 0`.
If the angle of inclination of the slope is doubled to `2theta^@`, then the acceleration of the object down the slope, in `text(ms)^(-2)`, is
`ma = mg\ sintheta`
| `sintheta` | `= a/g` |
| `cos^2theta` | `= 1-(a^2)/(g^2)` |
| `costheta` | `= sqrt(1-(a^2)/(g^2))` |
`text(If incline angle) = 2theta:`
| `ma` | `= mgsin(2theta)` |
| `a` | `= g*2sinthetacostheta` |
| `= g *2* a/g sqrt(1-(a^2)/(g^2))` | |
| `= (2a)/g sqrt(g^2-a^2)` |
`=>\ B`
An object with mass \(m\) kilograms slides down a smooth inclined plane with velocity \( \underset{\sim}{v}(t)\), where \(t\) is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle \(\theta\) with the horizontal, as shown in the diagram. The normal reaction force is \(\underset{\sim}{R}\). The acceleration due to gravity is \(\underset{\sim}{g}\) and has magnitude \(g\). No other forces act on the object.
The vectors \(\underset{\sim}{i}\) and \( \underset{\sim}{j} \) are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.
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i.
\(\text{Resolving forces in}\ \underset{\sim}{j} \ \text{direction:} \)
\( {\underset{\sim}{F}}_\underset{\sim}{j} = \underset{\sim}{R} + m\underset{\sim}{g}\ \cos \theta = 0\ \ \text{(in equilibrium)} \)
\(\text{Resolving forces in}\ \underset{\sim}{i} \ \text{direction:} \)
\( {\underset{\sim}{F}}_\underset{\sim}{i} = -m\underset{\sim}{g} \ \sin \theta \ \ \ \text{(down slope)} \)
\(\therefore \text{Resultant force:}\ \ \underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i} \)
ii. \(\text{Using}\ \ \underset{\sim}{F}=m \underset{\sim}{a}: \)
| \(m \underset{\sim}{a}\) | \(=-mg\ \sin \theta \ \underset{\sim}{i} \) | |
| \(\underset{\sim}{a}\) | \(=-g\ \sin \theta \ \underset{\sim}{i} \) | |
| \(\underset{\sim}{v}\) | \(= \displaystyle \int \underset{\sim}{a}\ dt \) | |
| \(=-gt\ \sin \theta +c \) |
\(\text{When}\ \ t=0,\ \ \underset{\sim}{v}=0\ \ \Rightarrow \ \ c=0 \)
\(\therefore \underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)
