
#1
Feb2110, 12:19 PM

P: 17

Two ropes are connected to a steel cable that supports a hanging weight as shown. If the maximum tension either rope can sustain without breaking is 5000 N, determine the maximum mass m that the ropes can support. 2. Relevant equations Newton's equations #1, #3. 3. The attempt at a solution Let the left rope have tension T_{1}, and the right rope have tension T_{2}. So, the horizontal component of T_{1} is cos(60)*T_{1}, which must equal the horizontal component of T_{2}, which is cos(40)*T_{2}. So: cos(60)*T_{1} = cos(40)*T_{2} T_{1} = (cos(40) / cos(60)) *T_{2} T_{1} = 1.532*T_{2} So T_{1} has the greater tension. The problem states that the maximum tension for either T_{1} or T_{2} is 5000N. So, since T_{1} has the greatest tension, we give it a tension of 5000N. This means that T_{2} has a tension of 3266N. The weight of the block is equal to the sum of the vertical components of the two tensions. So: w = sin(60)*T_{1} + sin(40)*T_{2} So we have: w = sin(60)*5000N + sin(40)*3266N w = 5035N That's the answer I get. My book tells me the answer is 6400N. I'm not sure where I've gone wrong here. Nevermind all of this. As it turns out, I was just screwing up the final calculation. sin(60)*5000N + sin(40)*3266N doesn't equal 5035N, as I indicated. So, nothing to see here... 


Register to reply 
Related Discussions  
Two ropes holding a weight. System is in equilibrium, how do I find T3?  Introductory Physics Homework  2  
Tension Problem (two ropes + weight)  Introductory Physics Homework  7  
determine maximum allowable weight of a load supported by chains  Mechanical Engineering  4  
Determine the maximum allowable weight of a load supported by chains.  Engineering, Comp Sci, & Technology Homework  1  
Ropes, Tension, Weight, & Equilibrium  Introductory Physics Homework  1 