Image- seesaw with 3 masses (A,B,C) and pivot below board of seesaw. Board is at
ID: 1912553 • Letter: I
Question
Image- seesaw with 3 masses (A,B,C) and pivot below board of seesaw. Board is at a diagonal so that it makes a positive slope with the pivot. On the seesaw shown, mass A is 17 kg, mass B is 10 kg, and mass C is 4.0 kg. Mass A is 0.50 m from the pivot. Mass C is 0.75 m from the pivot. The seesaw is at an angle of 25 degrees from the horizontal. Calculate the torques of mass A and mass C about the pivot. Where should you place mass B in relation to the pivot for the system to be in rotational equilibrium (i.e. there is no net torque on the system).Explanation / Answer
I will assume that mass A is on one side, and mass C is on the other side, so one distance must be taken as negative. So we have: M_a=60 and R_a=-1, M_b=30 and R_b=?, M_c=10 and R_c=+3. Question one. Find T_a and T_b. (really T=tau). In general, T=RxF, for the vectors T, R, and F, and the vector cross product x. If you know the magnitude of R and F, and the angle z (too lazy to write "theta") between R and F, then the magnitude, T=RFsin(z). For both masses, the force the weight (the force due to gravity), F_g=W=mg (let g=-9.8m/s^2), and points straight down. The R vector is 30 degrees above the horizontal. Therefore the angle between W and R is either z_right=90+30=120, or z_left=90-30=60. The first value is on the right, and the second value is on the left, thus sin(z_left)=sin(z_right)=sqrt(3)/2. So we will just use either and call it "z". Thus, T_a=R_a*(-M_a*g)*sin(z) = 588*(sqrt(3)/2) = 294*sqrt(3) = 509.2 N*m. Note the two negatives cancel out. T_b=R_b*(-M_b*g)*sin(z) = -294*(sqrt(3)/2) = -147*sqrt(3) = -254.6 N*m. Question two: Find R_b such that T_total=T_a+T_b+T_c=0. First, T_b=-R_b*M_b*g*sin(z). Secondly, T_a+T_c=-T_b. Putting it together, -R_a*M_a*g*sin(z)-R_c*M_c*g*sin(z)=R_b… The g*sin(z) drops out, leaving us with, -R_a*M_a-R_c*M_c=R_b*M_b. Thus, R_b= -(R_a*M_a+R_c*M_c)/M_b = +1 meters. Thus it must be on the right, 1 meter from the pivot point, and 2 meters before mass C.