A uniform plank of length 2.00 m and mass 33.5 kg is supported by three ropes, a

Question

A uniform plank of length 2.00 m and mass 33.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 715-N person is d = 0.725 m from the left end. Need Help? In an isometric exercise a person places a hand on a scale and pushes vertically downward, keeping the forearm horizontal. This is possible because the triceps muscle applies an upward force M perpendicular to the arm, as the drawing indicates. The forearm weighs 23.0 N and has a center of gravity as indicated. The scale registers 95 N. Determine the magnitude of M. N A hydrogen molecule consists of two hydrogen atoms whose total mass is 5.3 times 10-26 kg and whose moment of inertia about an axis perpendicular to the line joining the two atoms, midway between them, is about 1.88 times 10-46 kg.m2. Estimate, from these data, the effective distance between the atoms. m

Explanation / Answer

4) decompose T1 to horizontal and vertical components T1cos(40) & T1sin(40)

equilibrium implies

Fy = 0 T2+T1sin40 = weight of (plank+man) = 715+33.5*9.8=1043.3 N.......(i)

Fx=0 T3=T1cos40

d=0.725 m

applying moment balance equation about the point the man is standing

T2*d+335*(1-d) = T1sin40*(2-d).........(ii)

solving (i) and (ii)

T1= 660.03 N

T2=619.04 N

T3=505.61 N

5)the weight 23 N acts through the cg while the reaction force of 95 N acts at the palm placed on the scale at a distance of 0.300 m from elbow as shown in the diagram and Mas shown

applying moment balance eqn about elbow joint

23*0.150 -95*0.300 + M*0.0250=0

M=1002 N

6)mass of each atom be m

then 2m=5.3*10^-26 kg; m=2.65*10^-26 kg

let r be distance between them

then moment of inertia= 2*(m*(r/2)^2)=0.5 mr^2 =1.86*10^-46

r=1.184*10^-10 m

r/2=0.592*10^-10 m dist of each hydrogen atom from center

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