A man drags a 90-kg crate across the floor at a constant velocity by pulling on

Question

A man drags a 90-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted 25 degree above the horizontal, and the strap is inclined 61degree above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap. m = 90kg angle crate = 25 degree angle rope = 61 degree Length of the crate = 0.9m Width of the crate = 0.4m (cos theta)(l sin theta) + mg(L/2cos theta - w/2 sin theta) - (T sin theta)(L cos theta) = 0 T(L cos theta sin theta) - (L cos theta sin theta) = mg (w/2 sin theta - L/2 cos theta) T = mg(w/2 sin theta - L/2 cos theta)/(L cos theta sin theta) - (L cos theta sin theta) =(9.0)(9.0)((0.4/2 sin 61 degree) - (0.9/2 cos 25 degree))/0.9(cos 61 degree sin 25 degree) - (cos 25 degree sin 61 degree) =-205.43/-0.524 =388.33 N

Explanation / Answer

The net torque i zero:

m g (L/2) cos(25) = T L sin(61 - 25)

Solve for T:

T = (m g cos(25))/(2 sin(61 - 25))

T = ((90)*(9.8)*cos(25))/(2 sin(61 - 25))

T = 680 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.