Question

(a) A body of mass 2kg is placed on a rough plane inclined at an angle of 30° to the

Answer 1

Answer:

8.6 N

Step-by-step explanation:

Let T be the unknown force

Let θ be the slope of the inclined plane to horizontal

Let φ be the angle of the force to the plane

Let μ be the coefficient of static friction

Let m be the mass

Let N be the Normal force of plane on mass

Let Ff be the friction force which will have a maximum at μN

Let g be gravity

Forces acting parallel to the plane. Upslope is positive

For minimum force T, friction will be max and acting upslope.

                               F = ma

Tcosφ - mgsinθ + Ff = m(0)

Tcosφ - mgsinθ + μN = 0

Tcosφ - mgsinθ + μ(mgcosθ - Tsin(θ + φ)) = 0

T(cosφ - μsin(θ + φ)) + μmgcosθ - mgsinθ = 0

T  = mg(sinθ - μcosθ) / (cosφ - μsin(θ + φ))

T  = 2(9.8)(sin30 - 0.25cos30) / (cos30 - 0.25sin(30 + 30))

T = 8.5547537...