A 3.0 m long rigid beam with a mass of 110 kg is supported at each end. An 80 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?
Diagram: http://i137.photobucket.com/albums/q208/infinitbelt/p13-56.gif
I tried doing T = -(190 kg)(9.8)(-0.5 m) = 931 but that appears to be the wrong answers for support 2.
Any ideas?
Thanks!
2006-12-02 08:08:40 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The weight of the beam itself is evenly divided over the 2 supports.
The weight of the student however is going for 2/3 to support 2 and for 1/3 to support 1.
Hence support 1 exerts (9.81m/s^2)(110kg/2+80kg/3) = 801 N.
Support 2 exerts (9.81m/s^2)(110kg/2+80kg*2/3) = 1063 N.
Together this is 1864 N (=(110+80)*9.81)
2006-12-02 10:48:53 · answer #1 · answered by Frits 2 · 0⤊ 0⤋
Let the left support be considered a pivot and set the sum of torques = 0. By right-hand-rule, let CCW be positive.
- 110 kg*9.8 m/s^2*1.5 m - 80 kg*9.8 m/s^2*2 m + Fr*3 m = 0
where Fr is the upward force at the right support.
Fr = (1617 Nm + 1568) Nm/3 m = 1062 N
The total weight is 190 kg*9.8 m/s^2 = 1862 N. The left support provides the remainder of the support needed to hold up the load: 1862 N - 1062N = 800 N
2006-12-02 18:51:23 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋