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Salsk061 [2.6K]
3 years ago
6

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

result and then evaluate your answer if F = 260 N, b = 580 mm, and h = 370 mm. The moment is positive if counterclockwise, negative if clockwise.

Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

O=(0,0)

A=(0,-b)

B=(h,0)

We need to calculate the position vector of AB

\bar{AB}=(h-0)i+(0-(-b))j

\bar{AB}=hi+bj

We need to calculate the unit vector along AB

u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}

u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}

We need to calculate the force acting along the edge

\hat{F}=F(u_{AB})

\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})

We need to calculate the net moment

\hat{M}=\hat{F}\times OA

Put the value into the formula

\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})

\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))

\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})

\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}

Put the value into the formula

\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}

\hat{M}=-81.102\ \hat{k}\ N-m

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

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