- #1
paulius005
- 1
- 0
Block of mass M between two strings on incline?
A block of mass M rests on a frictionless inclined plane of angle θ as shown in the diagram below. Two springs of equal length are connected to the block and to two posts as shown. The separation between the posts is equal to the sum of relaxed lengths of the springs and the length of the block.
Suppose the top spring is stretched from its relaxed length by an amount δx = 0.082 m and the bottom string is compressed by the same amount.. If the spring constant of the top spring is twice that of the bottom spring and M = 0.25 kg and θ = 30o, what is k, the spring constant of the bottom spring?
http://i43.tinypic.com/2z4bcaw.gif
link to image
Now I know that I may need to use F = -k(X-Xo) and draw a free body diagram.
After drawing the free body diagram I get that mgsin(theta)=3k since there is one spring with spring constant k and another with 2k. I am not sure how to put the stretched length into all that though. Or since -k(X-Xo) = the vector some of the forces does that mean that mgsin(theta) = 3(-k(X-Xo)). Just a bit confused.
A block of mass M rests on a frictionless inclined plane of angle θ as shown in the diagram below. Two springs of equal length are connected to the block and to two posts as shown. The separation between the posts is equal to the sum of relaxed lengths of the springs and the length of the block.
Suppose the top spring is stretched from its relaxed length by an amount δx = 0.082 m and the bottom string is compressed by the same amount.. If the spring constant of the top spring is twice that of the bottom spring and M = 0.25 kg and θ = 30o, what is k, the spring constant of the bottom spring?
http://i43.tinypic.com/2z4bcaw.gif
link to image
Now I know that I may need to use F = -k(X-Xo) and draw a free body diagram.
After drawing the free body diagram I get that mgsin(theta)=3k since there is one spring with spring constant k and another with 2k. I am not sure how to put the stretched length into all that though. Or since -k(X-Xo) = the vector some of the forces does that mean that mgsin(theta) = 3(-k(X-Xo)). Just a bit confused.