Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Physics problem f=ma friction incline help?
Block M = 7:50 kg is initially moving up the incline and is increasing speed with
a = 4.87 m/s2. The applied force F is horizontal. The coefficients of friction
between the block and incline are Mu-s = 0:443 and Mu-k = 0:312. The angle of the incline is
25.0 degrees.
(a) What is the magnitude of the force F?
I've tried this one a lot but I can't calculate the normal force. Without the normal force i can't find the total friction and thereby cannot find the force f. Help?
because the force applied is horizontal on the block, it is pushing the block down which increases the normal force, so it isn't just mgsin(theta)
2 Answers
- FiremanLv 71 decade agoFavorite Answer
By F(net) = F(applied) - F(gravity) - F(friction)
=>m x a = F x cos 25* - mgsin25* - µk x N
=>7.5 x 4.87 = 0.91F - 7.5 x 9.8 x 0.42 - 0.312 x [mgcos25* + Fsin25*]
=>36.53 = 0.91F - 30.87 - 0.312 x 7.5 x 9.8 x 0.91 - 0.312 x 0.42F
=>F = 113.32 Newton
- 1 decade ago
I think what you are looking for in this problem is force of gravity in the y-direction. So I think it's 9.81sin(theta)
