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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.

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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.

Hence,

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.

• 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°

• 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

• Anonymous
1 month ago

North = 90° or 0° or 270°? THINK

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