Question

119 Part (6) Enter an expression for the net force in the horizontal direction. F. Your expression may include the normal force, FN, the applied force, F. the angle of the applied force above the horizontal, 0, and appropriate coefficients of friction. The friction coefficients are HR and 1196 Part (0) Enter an expression for the net force in the vertical direction. Your expression may include the mass, m addeleration due gravity.g, the normal force, Fx, the applied force, F, the angle of the applied force above the horizontal, 6, and appropriate coefficients of friction friction coefficients are pk and us A 11% Part (d) Enter an expression for the magnitude of the the normal force, Fx Your expression may include the mass, , the acceleration to gravity, g, the applied force, F, and the angle of the applied force above the horizontal, . 11% Part (0) Enter an expression for the magnitude of the applied force, F, at the threshold where static friction is at its maximum value. You expression may include the mass, m, the acceleration due to gravity g, the angle of the applied force above the horizontal, e, and appropriate coefficients of friction. The friction coefficients are us and us. 119 Part (1) Obtain a numeric value, in newtons, for the magnitude of the maximum applied force, F., consistent with static friction when the force makes an angle 29° above the horizontal and the mass of the block is 18 kg. 119 Part() Please use the interactive area below to draw a free body diagram to represent the situation where the contestant is pulling, and the block has just begun to accelerate. Use F, for the force of static friction and for the force of kinetic friction if they are needed for your free body 11. Part (h) Obtain an expression for the acceleration of the block corresponding to the free body diagram from part (g). Your expression may include the mass m, the acceleration due to gravity g, the applied force, F. the angle of the applied force above the horizontal, e, and appropriate coefficients of friction. The friction coefficients are and Pa O btainm ent value for the acceleration in meters per squared seconds, when the mass or the block is 8 ke and the angle or the rope is 29 above the horizontal. For the magnitude of the applied force, use the threshold value obtained in part

can you do all the problems please ?

Answer 1

13 Part - (a) Frino to Ecoso mg & free body diagram Past - 6 Ef as, & Coso - FS = Floso - MS - Flaso – us (mg F sino) Fit Fsing-mg N = mg - Fsing {f=&ese - es fong- & sind) Past (1) Ez = (FN + Fsino) - mg Part-(d) Normal foru, F = mg- & sing part (e) . At the threshold name of static friction, Floso = Fs ..

- Ford - Ms for & cost = us (ang - Fsind) & (Cost & as sing) = using 7 F & using (woso & M₂ Sino) (f) Giren, 7 = 22° m = 46 kg Ms = 0.03 The maximum applied force against the Static friction is As mg F = Cost & Ms sind 0.03X 4 689.8 cos 22 & 0.03 x Sin 22° = 14.41 Newton Frino 1 → Ecoso free body diagram -

acceleration of the block con se calculated as. ma = First - M (mg- I sino) a > a = F Cost - uk (mg- & sing) im ( Given, m = 46 kg o=220 Mr = 0.01 F = 14.41 & Coso - MK (ng. Fsind) ' a = .m 14.41 x cos22 - 0.01(4689.8 - 14.41 x Siner) 46 = 0.193 mest
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