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avanturin [10]
3 years ago
15

Determine the magnitude of the resultant force and the moment about the origin. Note: the symbol near the 140 N-m moment are not

additional forces. They represent the rotation caused with a magnitude of 140 N-m.

Engineering
1 answer:
arlik [135]3 years ago
7 0

Answer:

R = 148.346 N

M₀ = - 237.2792 N-m

Explanation:

Point O is selected as a convenient reference point for the force-couple system which is to represent the given system

We can apply

∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N

∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N

Then

R = √(Rx²+Ry²)    ⇒  R = √((66.8556 N)²+(132.4264 N)²)

⇒  R = 148.346 N

Now, we obtain the moment about the origin as follows

M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)

We can see the pic shown in order to understand the question.

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Hope that helps!

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3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
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Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

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