I NEED (E) PLEASE Problem 3 (40 points) The 150-lb man lies against the inclined

Question

I NEED (E) PLEASE

Problem 3 (40 points) The 150-lb man lies against the inclined cushion for which the coefficient of static friction is us 0.5. The man and the cushion are rotating about the z axis with a constant speed u = 20 ft/s. 10 8 ft farDetermine the normal and tangential accelerations the man is experiencing if he does not slide up or down along the incline (b) Determine the components of normal acceleration that are perpendicular to and parallel to the incline. teff 60°, determine the normal force the incline exerts on the man. dHf = 60°, determine the frictional forces the incline exerts on the man. (Assume friction acts up and right.) (e) Will static friction keep the man from sliding along the incline?

Explanation / Answer

Answer to part E of the question

From Newtons second Law

Alzebric sum of forces about axis n

Fn = man

0.5N cos + N sin = 150/32.2 (202/8)

N (0.5 cos + sin ) =150/32.2 *(202/8)

Fb = -150 + N cos -0.5N sin =0

N = 150 / (cos -0.5 sin)

Substiute N in abov equation we will get

150(0.5 cos + sin ) / (cos -0.5 sin) =150/32.2 (202/8)

(0.5 cos + sin ) = 1.553cos - 0.7764sin

1.7764sin = 1.053cos

tan = 1.053/1.7764

= 30.660

Here is the smallest angle of cushion at which the man will begin to slip off.

In Given problem the angle of static friction tan = = 0.5

= tan-1(0.5) =26.560

So the angle of static Friciton 26.560 is less than that of smallest cushion angle = 30.660 at which man starts sliding so static Friction keeps the man from sliding along the Incline.

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