<IMG alt=\"\" src=\"https://s3.amazonaws.com/answer-board-image/0575e7a2-f4d5-43

Question

<IMG alt="" src="https://s3.amazonaws.com/answer-board-image/0575e7a2-f4d5-4363-85c1-fe60c0917ca1.gif" src_cetemp="https://s3.amazonaws.com/answer-board-image/0575e7a2-f4d5-4363-85c1-fe60c0917ca1.gif">Two window washers, Bob and Joe, are on a 3.00 m long, 370 N scaffold supported by two cables attached to its ends. Bob weighs 805 N and stands 1.00 m from the left end, as shown in the figure below. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 980 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?<BR>left cable 1 kN <BR>right cable 2 kN <BR>

Explanation / Answer

T = 0 around the cable 0 = (1.5 m)(370 N) + (805 N)(1 m) + (500 N)(2 m) + (980 N)(2.5 m) + X(3 m) Where X = force the right cable exerts Solve for X, X = -1603.3, the negative sign means it is up. For the left cable: 0 = F + X + 370 + 805+500+980 F = -1603.3 + 370+805+500+980 = 1051.7 So: Right cable: 1603.3; Left cable: 1051.7

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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