@zareen said in MTH304 Assignment 1 Solution and Discussion:

A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10

Free Body Diagram

T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, Fs is the force of friction

Forces and their components on the x-y system of axis.

Equilibrium: W + T + N + Fs = 0

Forces represented by their components

W = (Wx , Wy) = ( - M g sin(35°) , - M g cos(35°))

T = (Tx , Ty) = (|T| cos (25°) , |T| sin (25°) )

N = (0 , Ny) = (0 , |N|)

Fs = (- |Fs| , 0) = ( - μs |N| , 0) , where μs is the coefficient of friction between the box and the inclined plane.

Sum of x components = 0

- M g sin(35°) + |T| cos (25°) + 0 - μs |N| = 0

which may be rewritten as

|T| cos (25°) = μs |N| + M g sin(35°)

sum of y components = 0
- M g cos(35°) + |T| sin (25°) + |N| + 0 = 0

|T| sin (25°) = M g cos(35°) - |N|

We now need to solve the system of two equations with two unknowns |T| and |N|.

|T| cos (25°) = μs |N| + M g sin(35°) (equation 1)

|T| sin (25°) = M g cos(35°) - |N| (equation 2)

solve equation 2 above for |N| to get

|N| = M g cos(35°) - |T| sin (25°)

Substitute |N| by M g cos(35°) - |T| sin (25°) in eq 1 to get

|T| cos (25°) = μs [ M g cos(35°) - |T| sin (25°) ] + M g sin(35°)

rewrite above equation as follows

|T| [ cos (25°) + μs sin (25°) ] = μs M g cos(35°) + M g sin(35°)

Solve for |T|

|T| =

μs M g cos(35°) + M g sin(35°)

cos (25°) + μs sin (25°)

Substitute with numerical Values

μs = 0.3, M = 10 Kg, g = 10 m/s^2

|T| ≈ 79.3 N

Use |N| = M g cos(35°) - |T| sin (25°) found above

|N| = 100 cos(35°) - 79.3 sin (25°) ≈ 48.4 N