(a) A body of mass 2kg is placed on a rough plane inclined at an angle of 30° to the horizontal. The coefficient of friction between the body and the plane is 0.25. Find the
least force needed to prevent the body from slipping down the plane if this force acts
upwards at an angle of 30° to the line of greatest slope.

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...

step-by-step explanation:

i hate black

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0