3,00 2,00 1,00 0,00 0,00 0,20 0,40 0,60 0,80 1,00 1/m2 (kg) 1,20 x(m) 4,00 y=1.995x+1,504...

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3,00 2,00 1,00 0,00 0,00 0,20 0,40 0,60 0,80 1,00 1/m2 (kg) 1,20 x(m) 4,00 y=1.995x+1,504 3,50 3,00 2,50 2,00 1,50 1,00 0,50 0,00 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 3 2,5 2 1,5 1 0,5 1,396x+1,052 0 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 Part C Choose the correct plot of versus 1/m2 with the correct equation of the best-line fit. x(m) 3,50 y=1.995x+0,974 3,00 2,50 2,00 1,50 1,00 0,50 0,00 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 6,00 5,00 4,00 3,00 y=2.993x+2,256 2,00 1,00 0,00 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 4.00 1/m2 (kg) Act Go t Part B Apply the static equilibrium equation T = 0 with the axis at the location of the knife-edge pivot. Solve the equation for as a function of m2 (2.60 0.38M); +1.50 m x = m2 1 x = (2.60 0.38M).. -1.30 m m2 x = (2.60 0.38M). +1.30 m m2 Ox = (2.60+0.38M).: -1.50 m m2 Submit Request Answer You need to measure the mass M of a 4.00-m-long bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn't in the middle of the bar. The bar is too long for you to weigh on your scale. So, first you balance the bar on a knife-edge pivot and determine that the bar's center of gravity is 1.88 m from its left-hand end. You then place the bar on the pivot so that the point of support is 1.50 m from the left-hand end of the bar. Next you suspend a 2.00 kg mass (m1) from the bar at a point 0.200 m from the left- hand end. Finally, you suspend a mass m = 1.00 kg from the bar at a distance x from the left-hand end and adjust so that the bar is balanced. You repeat this step for other values of m2 and record each corresponding value of x. The table below gives your results. m2 (kg) 1.00 1.50 2.00 2.50 3.00 4.00 x (m) 3.50 2.83 2.50 2.32 2.16 2.00 Figure cg m2 < 1 of 1 > Part A Draw a free-body diagram for the bar in (Figure 1) when m and m2 are suspended from it. Draw the vectors starting at the black dots. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. + No elements selected m F cg W m2 i C 3,00 2,00 1,00 0,00 0,00 0,20 0,40 0,60 0,80 1,00 1/m2 (kg) 1,20 x(m) 4,00 y=1.995x+1,504 3,50 3,00 2,50 2,00 1,50 1,00 0,50 0,00 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 3 2,5 2 1,5 1 0,5 1,396x+1,052 0 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 Part C Choose the correct plot of versus 1/m2 with the correct equation of the best-line fit. x(m) 3,50 y=1.995x+0,974 3,00 2,50 2,00 1,50 1,00 0,50 0,00 1/m2 (kg) 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 6,00 5,00 4,00 3,00 y=2.993x+2,256 2,00 1,00 0,00 0,00 0,20 0,40 0,60 0,80 1,00 1,20 x(m) 4.00 1/m2 (kg) Act Go t Part B Apply the static equilibrium equation T = 0 with the axis at the location of the knife-edge pivot. Solve the equation for as a function of m2 (2.60 0.38M); +1.50 m x = m2 1 x = (2.60 0.38M).. -1.30 m m2 x = (2.60 0.38M). +1.30 m m2 Ox = (2.60+0.38M).: -1.50 m m2 Submit Request Answer You need to measure the mass M of a 4.00-m-long bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn't in the middle of the bar. The bar is too long for you to weigh on your scale. So, first you balance the bar on a knife-edge pivot and determine that the bar's center of gravity is 1.88 m from its left-hand end. You then place the bar on the pivot so that the point of support is 1.50 m from the left-hand end of the bar. Next you suspend a 2.00 kg mass (m1) from the bar at a point 0.200 m from the left- hand end. Finally, you suspend a mass m = 1.00 kg from the bar at a distance x from the left-hand end and adjust so that the bar is balanced. You repeat this step for other values of m2 and record each corresponding value of x. The table below gives your results. m2 (kg) 1.00 1.50 2.00 2.50 3.00 4.00 x (m) 3.50 2.83 2.50 2.32 2.16 2.00 Figure cg m2 < 1 of 1 > Part A Draw a free-body diagram for the bar in (Figure 1) when m and m2 are suspended from it. Draw the vectors starting at the black dots. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. + No elements selected m F cg W m2 i C