8.6 Analysis framework for laws of motion  (Page 3/3)

The real time situation presents real three dimensional bodies. We encounter situation in which forces are not concurrent. Still, we consider them to be concurrent (by shifting force in parallel direction). It is because physical set up constrains motion to be translational. For example, consider the motion of a block on an incline as shown in the figure.

A block on an incline

The forces weight (mg), normal force (N) and friction force (Ff) are not concurrent. The friction force, as a matter of fact, does not pass through "center of mass" and as such induces "turning" tendency. However, block is not turned over as restoring force due to the weight of the block is greater and inhibits turning of the block. In the nutshell, block is constrained to translate without rotation. Here, translation is enforced not because forces are concurrent, but because physical situation constrains the motion to be translational.

Resolving force along the coordinate axes

The application of a force and resulting motion, in general, is three dimensional. It becomes convenient to analyze force (cause) and acceleration (effect) analysis along coordinate axes. This approach has the benefit that we get as many equations as many there are axes involved. In turn, we are able to solve equations for as many unknowns.

We must know that the consideration in each direction is an independent consideration – not depending on the motion in other perpendicular directions. A force is represented by an equivalent system of components in mutually perpendicular coordinate directions.

The component of force vector along a direction (say x-axis) is obtained by :

Component of force

where “θ” is the angle that force vector makes with the positive direction of x-axis. For example, consider the force as shown in the figure. We need to find the component of a force along x-axis. Here,

Component of force

$$F_x=F\cos \theta =10\cos {240}^0=10X-\frac{1}{2}=-5N$$

Alternatively, we consider only the acute angle that the force makes with x-axis. We need not measure angle from the The angle here is 60°. As such, the component of the force along x-axis is :

Component of force

$$F_x=F\cos \theta =10\cos {60}^0=10X\frac{1}{2}=5N$$

We decide the sign of the component by observing whether the projection of the force is in the direction of x-axis or opposite to it. In this case, it is in the opposite direction. Hence, we apply negative sign as,

$$F_x=-F\cos \theta =-10\cos {60}^0=-10X\frac{1}{2}=-5N$$

Clearly, if a force “ F ” makes angles α, β and γ with the three mutually perpendicular axes, then :

Components of force

$$F_x=F\cos \alpha$$

$$F_y=F\cos \beta$$

$$F_z=F\cos \gamma$$

We can write vector force, using components, as :

In the case of coplanar force, we may consider only one angle i.e. the angle that force makes with one of the coordinate direction. The other angle is compliment of this angle. If “θ” be the angle that force makes with the positive direction of x-axis, then

Components of coplanar force

$$F_x=F\cos \theta$$

$$F_y=F\cos \left({90}^0-\theta \right)=F\sin \theta$$

Problem : Find the magnitude of components of the weight of the block of 1 kg in the directions, which are parallel and perpendicular to the incline of angle 30°.

Solution :

The weight of the block acts in vertically downward direction. Its magnitude is given as :

Components of force

$$W=mg=1X10=10N$$

The direction of weight is at an angle 30° with the perpendicular to the incline. Hence, components are :

$$W_{\perp }=W\cos \theta =10\cos {30}^0=10X\frac{\sqrt{3}}{2}=5\sqrt{3}N$$

$$W_{\mid \mid }=W\sin \theta =10\sin {30}^0=10X\frac{1}{2}=5N$$

Note : We should realize here that these are the magnitudes of components in the referred directions. The components have directions as well. The values here, therefore, represent the scalar components of weight. The vector components should also involve directions. Since components are considered along a given direction, it is possible to indicate direction by assigning appropriate sign before the values.

The sign of the components depend on the reference directions i.e. coordinate system. Let us consider the given components in two coordinate systems as shown. In the first coordinate system,

Components of weight

$$W_y=-W\cos \theta =-10\cos {30}^0=-10X\frac{\sqrt{3}}{2}=-5\sqrt{3}N$$

$$W_x=W\sin \theta =10\sin {30}^0=10X\frac{1}{2}=5N$$

In the second coordinate system,

$$W_y=W\cos \theta =10\cos {30}^0=10X\frac{\sqrt{3}}{2}=5\sqrt{3}N$$

$$W_x=W\sin \theta =10\sin {30}^0=10X\frac{1}{2}=5N$$

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