R is the NormalReaction. P is Frictionalforce. 0.3 is coefficient ofstatic frict

Question

R is the NormalReaction.

P is Frictionalforce.

0.3 is coefficient ofstatic friction (Assumed).

The Frictional force P will act upwards, as the body is at thepoint of sliding downwards, now resolving

the forces along the plane, assuming the coefficient of staticfriction (u)=0.3

     F*cos20=70*sin35-0.3*R                  ---------------- (1)

       And now resolving theforces perpendicular to the plane,

R=70*cos35-F*sin20                   ---------------- (2)

Now substituting inequation (1) we get Reaction R as

0.3*R=(70*sin35)-(27.41*cos20)

R=48 lb

Frictional ForceP=0.3*48=14.4 lb

Explanation / Answer

the solution one is wrong, the friction force f should be up theincline. let gravity w = 70 lb, = 35o, =20o, = 0.30 Fcos + f - wsin = 0, so f = wsin -Fcos N + Fsin - wcos = 0, so N = wcos- Fsin use f = N wsin - Fcos = (wcos - Fsin) F = w*(sin - cos)/(cos -sin) = 27.4 lb

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