PROBLEM 3: A staging that weighs 290 N supports two painters, one 430 N and the other 490 N. The reading in the left scale is Fl = 500 N. What is the reading Fr in the right hand
scale? Further description is the picture in the pdf.
PROBLEM 4:So basically in problem 4, there is a 56 N weight suspended in the air by two ropes. One of the ropes is slanted at 53 degrees to the ceiling, and the other rope is perpendicular to the wall. Both intersect and join at a point and then descend vertically as one rope attached to the weight. If you need any more visual descriptions, a picture is included in the pdf.
PROBLEM 3: soh cah toa (sin=opposite/hypotenuse, cos=adjacent/hypotenuse, and tangent=opposite/adjacent)
PROBLEM 4: soh cah toa (sin=opposite/hypotenuse, cos=adjacent/hypotenuse, and tangent=opposite/adjacent)
The Attempt at a Solution
PROBLEM 3: So I knew the total downward force of the staging and the painters combined: 430 N+490 N+290 N= 1210 N. And if the reading on the scale was 500 N on the left side (assuming 430 N is also on the left side), then that means that the upward force from the right rope was 220 N. I don't know how to solve this.
PROBLEM 4: I basically extended the vertical rope until it hit the ceiling, so the section between the horizontal and the diagonal ropes was split into a right triangle and a rectangle. I figured that the theta of the right triangle was 37 (90-53) and then used sin (37) to get 0.6018150231520482799179770004414898414256377. This gave me the ratio of two lengths that I didn't know. I then moved on to find the forces. So if the force on the block was 56 N, that mean that the adjacent of the right triangle would have to be equal to that because the block was in equilibrium. Now i just don't know what to do next.
Thank you for reading! Oh, please help, by the way.
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