Question

A)

You pull on a large box using a rope as in (Figure 1), except the rope is at an angle of 15.0∘ below the horizontal. The weight of the box is 325 N , and the coefficient of kinetic friction between the box and the floor is 0.270. What must be the tension in the rope to make the box move at a constant velocity?

Express your answer with the appropriate units..

B) What is the normal force that the floor exerts on the crate?

Express your answer with the appropriate units.

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**** I will add more parts in the comments after you solve those two parts so I could give you a good rate ^_^ .. it will open after I answer those two

Answer 1

(A) The free body diagram is given as follows: t7sin150 150 Tcos150 Here, fis the frictional force, Tis the tension in the rope, and R is the surface reaction. ΣΕ-0 T sin 15°+ R-W = 0 T sin 15°+ R-325 N =0 So, R = 325 N-T sin 15° (1) Here, the velocity is constant. Thus, ΣΕ-0 T cos 15°- f =0 T cos15°- 4,R =0 I cos 15° -(0.270)R = 0 Substitute equation (1) for R. T cos 15° -(0.270)(325 N-T sin 15°) = 0 Tcos15° +(0.270) T sin 15° = (0.270)(325 N) T(cos15°+(0.270) sin 15°) = (0.270)(325 N) (0.270)(325 N) T = (cos 15° +(0.270) sin15°) =84.7 N (B) The normal force can be calculated using equation (1). R = 325 N-T sin 15° Substitute 84.7N for Tin the above equation. R = 325 N-(84.7 N) sin 15° = 303 N