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sreya
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Homework Statement
A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Figure 1) .
Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 25.0N , F2 = 15.5N , and F3 = 17.0N . The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise.
Homework Equations
[itex]\tau=rFsin(\phi)[/itex]
The Attempt at a Solution
[itex]rF2+rF3-rF1*sin(45) = .675[/itex], r=.09
Apparently that's incorrect though, could anyone provide insight as to why?
Solution
There are two ways to express torque:
[itex]\tau = r(Fsin\phi)=rF_t[/itex]
[itex]\tau = rsin\phi F =r_{\perp}F[/itex]
So, the solution would go like so:
[itex]r_1=r_2=r_3=\frac{l}{2sin\theta}[/itex] where [itex]\theta=45[/itex] and l = length of side
[itex]r(F_1*sin(45)+F_3-F_2*sin(45))[/itex]
Since the moment arm ([itex]t_{\perp}[/itex]) forms a 45 degree angle with F1 and F2
So unlike i tried in the beginning the real equation has the form:
[itex]F_1r_{\perp}-F_2r_{\perp}+rF_t[/itex] where [itex]F_3=F_t[/itex]
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