Two blocks are positioned on surfaces, each inclined at the same angle q = B) 23
ID: 1553441 • Letter: T
Question
Two blocks are positioned on surfaces, each inclined at the same angle q = B) 23.8 _o with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of the black block is C) 4.7 kg, and the coefficient of friction for both blocks and inclines is m = D) 0.07. Assume gravity is g = 9.8 m/s2 .
A. What is must be the mass of the white block if both blocks are to slide to the right at a constant velocity? Answer: ___.___ kg
B. What is must be the mass of the white block if both blocks are to slide to the left at a constant velocity? Answer: ____.___ kg
C. What is must be the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5 m/s2? Answer: ____.___ kg
D. What is must be the mass of the white block if both blocks are to slide to the left at an acceleration of 1.5 m/s2? Answer: ___.___ kg
E. Now, re-do part (d) above, but now assuming no friction at all on either incline. So in the absence of friction, what must be the mass of the white block such that both blocks slide to the left at an acceleration of 1.5 m/s2? Answer: ________.___ kg
Explanation / Answer
a) tension in string for black block
mgsin23.8 = T + 0.07*mg*cos23.8
4.7*9.8*sin23.8 = T + 0.07*4.7*9.8*cos23.8
T = 15.64 N
for white block
T = 0.07Mw*gcos23.8 +Mw*gsin23.8
15.64 = 0.07*Mw*9.8*cos23.8 +Mw*9.8*sin23.8
Mw = 3.41 kg
2)in this case
tension in string for black block
T = 0.07*mg*cos23.8 + mgsin23.8
T = 0.07*4.7*9.8*cos23.8 + 4.7*9.8*sin23.8
T = 21.54
for white block
mgsin23.8 = T + 0.07* mgcos23.8
m*9.8*sin23.8 = 21.54 + 0.07* m*9.8*cos23.8
m = 6.47 kg
3)in this case
tension in string for black block
mgsin23.8 - T - 0.07*mgcos23.8 = ma
4.7*9.8*sin23.8 - T - 0.07*4.7*9.8*cos23.8 = 4.7*1.5
T = 8.587 N
for white block
T - (0.07*Mw*gcos23.8 +Mw*gsin23.8) = Mw*1.5
8.587 - (0.07*Mw*9.8*cos23.8 +Mw*9.8*sin23.8) = Mw*1.5
Mw = 1.41 kg
4)in this case
tension in string for black block
T - (0.07*mg*cos23.8 + mgsin23.8) = m*1.5
T - (0.07*4.7*9.8*cos23.8 + 4.7*9.8*sin23.8) = 4.7*1.5
T = 28.587
for white block
mgsin65 - (T + 0.07* mgcos65) = m*1.5
m*9.8*sin23.8 - (28.587 + 0.07* m**9.8*cos23.8) = m*1.5
m = 15.65 kg
5) tension in string for black block
T - mgsin23.8 = m*1.5
T - 4.7*9.8*sin23.8 = 4.7*1.5
T = 25.64 N
for white block
mgsin23.8 - T = m*1.5
m*9.8*sin23.8 - 25.64 = m*1.5
m = 10.45 kg
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.