Question

My question is ... and I m not understanding. Force along side = W cos theta. how it came. please help me out with porper explanation please please​

Answer 1

GIVEN :-

• Weight of the block = 50N

• Inclination = 30°

TO FIND :-

• Magnitide of the Force acting along the slide

SOLUTION :-

First of all Let's draw Free body diagram (FBD) for the given system. The forces acting on the body are

• weight (mg verticelly downward)

• Normal ( Contact force acting perpendicular to the surface of contact )

By splitting mg into two components we get mgSinθ and mgCosθ

From the diagram we can say that ,

The magnitude of Force acting down along the slide is equal to mgSinθ

We have ,

• W = mg = 50N

• θ = 30°

${\implies {\sf \: F \: = mg \sin( \theta)}} \\ \\ {\implies \:{\sf F = 50 \times \sin(30)}}$

$\large {\bold {\boxed{ \sf{ \sin(30) = \dfrac{1}{2} }}}}$

$\implies \:\sf\: F = 50 \times \dfrac{1}{2} \\ \\ \implies \: F = 25 \:N$

∴ The magnitide of Force acting downward along the inclined plane is 25 N

NOTE :-

• While splitting the given vector x (let) into two components We decide the certain component as xcosθ if the component is adjacent to θ and the component which is opposite i.e not adjacent to θ such component is decided as xSinθ and these components must be perpendicular to each other giving the resultant as mg

• The Angle between mg and MgCosθ is θ because the angles , inclination of inclined plane and the angle between mg and mgCosθ is equal since they are Alternate interior angles (Refer the image)