A 30 kg box is placed on a 15? slope as shown below, and a person pushes on the

Question

A 30 kg box is placed on a 15? slope as shown below, and a
person pushes on the box up the slope to keep it from sliding. If
the coefficient of static friction is 0.15, what is the minimum
force that the person must exert to keep the box in place?

A 30 kg box is placed on a 15? slope as shown below, and a person pushes on the box up the slope to keep it from sliding. If the coefficient of static friction is 0.15, what is the minimum force that the person must exert to keep the box in place? A. 273 N B. 241 N C. 33.5 N D. 64.7 N E. 294 N 3 masses are connected together with massless ropes and frictionless pulleys as shown below. The center mass weighs 150 N, and the weight of the other masses is unknown. Rope A is at the angle shown. Rope B is horizontal, and rope C is vertical. 5. What is the tension in rope C? A. 29 N B. 47 N C. 116 N D. 13 N E. 150 N 6. What is the tension in rope A? A. 47 N B. 75 N C. 116 N D. 150 N E. 173 N

Explanation / Answer

sum forces along the slope

F + friction - mg sin theta = 0

friction = u N = u m g cos theta

F = m g sin theta - um g cos theta

F = 30*9.81*(sin(15 degrees) - 0.15*cos(15 degrees))= 33.5 N

5) since it has to hold up 150 N, it is 150 N so E

6) so sum forces in the y

T sin 60 - 150 = 0

t = 150/sin(60 degreeS)= 173 N

so E)

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