JLTVI Tue wouy LYF WUUN JIULI TULIVIT VELIT TIL LALU wyprut TCTULUIT LV LTC LAJ VI TUULLIVET U pon. Point at which force

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Jltvi Tue Wouy Lyf Wuun Jiuli Tulivit Velit Til Lalu Wyprut Tctuluit Lv Ltc Laj Vi Tuullivet U Pon Point At Which Force 1 (120.38 KiB) Viewed 51 times

JLTVI Tue wouy LYF WUUN JIULI TULIVIT VELIT TIL LALU wyprut TCTULUIT LV LTC LAJ VI TUULLIVET U pon. Point at which force acts · Line of action of force Position vector from point to the point at which the force acts Angle between line of action of force and radial direction 0=109 11 From In our first example, we will see that we can use a lever arm to greatly increase the torque without having to provide more force. Suppose, for instance, that an amateur plumber, unable to loosen a pipe fitting, slips a piece of scrap pipe (sometimes called a "cheater") over the handle of his wrench. He then applies his full weight of 900 N to the end of the cheater by standing on it. The distance from the center of the fitting to the point where the weight acts is 0.80 m, and the wrench handle and cheater make an angle of 19° with the horizontal (Figure 1). Find the magnitude and direction of the torque of his weight about the center of the pipe fitting -0.80 m 1907 -- Point where axis of rotation intersects plane of diagram T is positive. F-900 NV ·Moment arm (perpendicular distance from axis of rotation to line of action of force) = SOLVE We have a choice of methods for solving this problem. From the free-body diagram, the angle o between the vectors † and Fis 109°, and the moment arm l is 1=(0.80 m) (sin 109 ) = (0.80 m) (sin 71°) = 0.76 m. We can find the magnitude of the torque from either T=Flort=rF sin . From T=FI, T=Fl= (900 N)(0.76 m) = 680 N.m. Or, from t=rFsin , t=rF sin = (0.80 m)(900 N) (sin 109") = 680 N.m. ALTERNATIVE SOLUTION Alternatively, we can start with the components of F. From the free-body diagram, we see that the component of force F perpendicular to (which we call Ftan) is Ftan = F(cos 19") = (900 N) (cos 19°) = 850 N. = Then the torque is T= Ftan? =(850 N)(0.80 m) = 680 N.m. REFLECT The force tends to produce a counterclockwise rotation about O, so with the convention described in your textbook, this torque is +680 N·m not -680 Nm. Also, please don't try to use this plumbing technique at home, you're very likely to break a pipe. Part A - Practice Problem: Figure < 1 of 1 1 If the pipe fitting under consideration can withstand a maximum torque of 650 N·m without breaking, what is the smallest angle at which our amateur plumber can safely stand on the end of the cheater and apply his full weight without breaking the pipe fitting? Express your answer in degrees. ΟΙ ΑΣφ ? 0.80 m a = a = O F = 900 N 19° Submit Request Answer