If you know a crate weighs 500 N, the coefficient of kinetic friction is 0.40 an

Question

If you know a crate weighs 500 N, the coefficient of kinetic friction is 0.40 and you know the crate is being pulled on a horizontal surface by a rope at an angle of 30 degrees at a constant velocity by a force "T". In component form F(x)= Tcos30-0.40n=0 and F(y)= Tsin30 + normal force + weight (-500 N)= 0
to solve for normal force you can use the 2nd equation (Fy) so you get n = 500N- Tsin30 and now you can substitute this into the first equation getting Tcos30- 0.40(500 N - Tsin30)=0 . You can solve for T and get 188 N for T and 406 N for normal force says my textbook, however I am having problems solving for T in the last equation can someone work it out step by step and show me, i am making a mistake in my calculations and not getting the right answer. thank you

Explanation / Answer

so we have Tcos30- 0.40(500 N - Tsin30)=0 multiply the 0.4 through T cos 30 - 200 N + 0.4 T sin 30 = 0 move the 200N to the other side Tcos 30 + 0.4 T sin 30 = 200 factor the T T (cos 30 + 0.4 sin 30) = 200 divide over T= 200/(cos30 + 0.4 sin30)=188 N plug that back into equation for n n= 500 -Tsin30 = 406 N

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