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Mechanics- forces in two dimensions ?
A ship of mass 10000 kg is being towed due north by two tugboats with acceleration 0.1m/s^2. One pulls with a tension of 2000N on a bearing of 330 degree. The other pulls with a tension, T on a bearing of theta. There is resistance against the motion of 1000 N. Find T and theta.
- az_lenderLv 71 month agoFavourite answer
The applied force is (ma + 1000N) northward,
which is 2000N northward.
So you have
(2000 N)(-i/2 + j*sqrt(3)/2) + T(i sin(theta) + j cos(theta)) =
= (2000 N) j.
T*sin(theta) = 1000 N and
T*cos(theta) = 2000 N - 1732 N = 268 N.
Then tan(theta) = 1000/268 = 3.731 and theta = 75 degrees, and T = 1035 N.
- oubaasLv 71 month ago
T1 = 2000 N
T1x = 2000*cos 30 = 1730 N
T1y = -2000*sin 30° = -1000 N
Tx = -T1x = -1730 N
Ty = -T1y + Fattr. + m*a = 1000+1000+10.000*0.1 = 3000 N
T = Tx^2+Ty^2 =
T = √((√3)/2)^2*2000+3000^2 = 1000√3/4*2^2+3^2 = 1000√3+9 = 2000√3 N
bearing angle Θ = 180+arctan -Tx/Ty = 180+arctan (√3)/3 = 180+30 = 210°
- billrussell42Lv 71 month ago
330º is 30º west of N.
The N component is 2000cos30 = 1732 N
The W component is 2000sin30 = 1000 N
net N force on the ship is sum of
F = ma = 10000•0.1 = 1000 N
added to 1000 N resistance = 2000 N
It is all N, no W or E component
net W or E force is zero
so N component of second tug is 2000–1732 = 268 N
and W component = –1000 N or E force of 1000 N
vector sum of the two is √(1000²+268²) = 1035 N
angle θ = arctan (268/1000) = 15.0º which would be, looking at the vectors, 75º E of N
- Anonymous1 month ago
North = 90° or 0° or 270°? THINK