Solve Tension on Wall-Hung Picture Frame Wire

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To solve the tension on a wall-hung picture frame with a 40-inch wire and 39-inch spaced eyebolts, it is essential to visualize the setup, as the wire forms a triangle. The weight of the picture frame is 8 pounds, which influences the tension in the wire. By drawing a diagram and labeling the forces and moments, one can apply static equilibrium principles to analyze the situation. The tension in the wire can be calculated using trigonometric relationships based on the triangle formed by the wire and the distance between the eyebolts. Understanding these concepts is crucial for accurately determining the tension in the wire.
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i need help on this question.
a picture is hung on a wall. the wire is 40in. long and the eyelits that the wire is connected to are 39in. apart. the weight of the picture frame is 8lbs. what is the tension on the wire?
 
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This is a statics problem. It is best to draw a picture - it doesn't have to be super accurate, but just show what's going on. Then label the forces and moments involved and sum them in each coordinate direction. Hint: the wire will make a triangle because it is longer than the eyelits are apart.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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