- #1
buffgilville
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1) A plank of mass 2.5 kg and length 2 meters is resting horizontally on two wedges. The first is at one of the ends and the other is a distance 0.3 meters from the second end. If a mass of 0.2 kg is placed on this second edge, what is the force (in Newtons) acting on the second wedge.
Here's what I did:
T1=F1(0) = 0
T2=F2(1.7)
T(mass)=-(0.2*9.81*2) = -3.924
T(plank)=-(2.5*9.81*(2/2)) = -24.525
sum of torque = F1(0) + F2(1.7) - 3.924 - 24.525 = 0
so, F2=16.73 Newtons
but the correct answer is 1.71 Newtons. Where did I go wrong?
2) The mass of the Earth is 6.0E24 kg and its radius is 3950 miles. Assuming that the Earth is a uniform sphere, its angular momentum (in Joule. Secs) is aE+33 where a is?
I = (2/5) MR^2 ---> (2/5)(6.0E24)(6355550meters)^2 = 9.694E37
L = Iw ---> (9.694E37)((2pi)/86400) = 7.0499E33
which makes a=7.050
but the correct answer is 6.970. (I did not round anything either.) What did I do wrong? Please help. Thanks!
Here's what I did:
T1=F1(0) = 0
T2=F2(1.7)
T(mass)=-(0.2*9.81*2) = -3.924
T(plank)=-(2.5*9.81*(2/2)) = -24.525
sum of torque = F1(0) + F2(1.7) - 3.924 - 24.525 = 0
so, F2=16.73 Newtons
but the correct answer is 1.71 Newtons. Where did I go wrong?
2) The mass of the Earth is 6.0E24 kg and its radius is 3950 miles. Assuming that the Earth is a uniform sphere, its angular momentum (in Joule. Secs) is aE+33 where a is?
I = (2/5) MR^2 ---> (2/5)(6.0E24)(6355550meters)^2 = 9.694E37
L = Iw ---> (9.694E37)((2pi)/86400) = 7.0499E33
which makes a=7.050
but the correct answer is 6.970. (I did not round anything either.) What did I do wrong? Please help. Thanks!
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