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Physics For Scientists & Engineers
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Short Answer

A uniform plank of length 2.00 m and mass 30.0 kg is supported by three ropes as indicated by the blue vectors in Figure P12.25. Find the tension in each rope when a 700-N person is d=0.500m from the left end.

The second rope has a tension of =672.1N

The third rope has a tension of =382.78N

Anticlockwise torque is considered good, whereas clockwise torque is considered negative.

See the step by step solution

Step by Step Solution

Step 1: Conceptualize:

Consider this: For any item to be in equilibrium, the net forces operating on it must be zero, as well as the net torque acting on it.

Step 2: Categorize:

Because the plank must be balanced,

The sum of upward and downward forces equals the sum of downward forces.

i.e. ΣFup =ΣFdimn

The torques around a point should add up to zero.

i.e. Στ=0

Step 3: Analyze:

The net vertical force is (from the free body diagram)

T2+T1sin40°=Mg+wT2+T1sin40°=(30 kg)(9.8 m/s2)+700 NT2+T1sin40°=994 N

The horizontal force is net.

T3=T1cos40°

In a counterclockwise route, take torque around the location where the man stood.

T2dMg(L2d)+T1sin40°(Ld)=0T2(0.5 m)147 N+T1sin40°(2 m0.5 m)=0T2=3T1sin40°294

Step 4: Determining the tension:

Using the equations above,

3T1sin40°294+T1sin40°=994 NT1sin40°=322 NT1=500.94 NT1=501 N

The second rope has a tension of

T2=3T1sin40°294=3(501)sin40°294=672.1 N

The third rope has a tension of

T3=T1cos40°=(501 N)cos40°=383.78 N

Step 5: Conclusion:

Anticlockwise torque is considered good, whereas clockwise torque is considered negative.

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