A horizontal uniform plank is supported by ropes I and II at points P and Q, respectively, as shown above. The two ropes have negligible mass. The tension in rope I is 150 N. The point at which rope II is attached to the plank is now moved to point R halfway between point Q and point C, the center of the plank. The plank remains horizontal. What are the new tensions in the two ropes?

The answer is T1=100N and T2=200N but I don't know the steps to solve this one. An explanation would be much appreciated.

The new tensions in the two ropes after the movement of rope 2 are;

T₁ = 100 N

T₁ = 100 NT₂ = 200 N

We are told that as the plank is currently, the two ropes attached at each end have tension of 150 N each.

Thus;

T₁ = T₂ = 150 N

The two ropes are acting in tension upwards and so for the plank to be balanced, there has to be a downward force(which is the weight of the plank) must be equal to the sum of the tension in the two ropes.

Thus, from equilibrium of forces, we have;

W = T₁ + T₂

W = 150 + 150

W = 300 N

Now, we are told that;

Rope 2 is now moved to a point R which is halfway between point C and Q. Since C is the centre of the plank and R is the midpoint of C and Q, if the length of the plank is L, then the distance of rope 2 from point P is now ¾L.

Since the plank remains horizontal after shifting the rope 2 to point R, let us take moments about point P to get;

T₂(¾L) - W(½L) = 0

Plugging in the relevant values;

T₂(¾L) - 300(½L) = 0

T₂(¾L) - 150L = 0

Rearrange to get;

T₂(¾L) = 150L

Divide both sides by L to get;

T₂(¾) = 150

Cross multiply to get;

T₂ = 150 × 4/3

T₂ = 200 N

Thus;

T₁ = 300 - 200

T₁ = 100 N

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Explanation:

There are three forces on the plank.  T₁ pulling up at point P, T₂ pulling up at point Q, and W pulling down at point C.

Let's say the length of the plank is L.

Sum of forces in the y direction before rope II is moved:

∑F = ma

150 N + 150 N − W = 0

W = 300 N

Sum of moments about point P after rope II is moved:

∑τ = Iα

(T₁) (0) − (300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (1/2) + (T₂) (3/4) = 0

-150 N + 3/4 T₂ = 0

T₂ = 200 N

Sum of forces in the y direction:

∑F = ma

T₁ + 200 N − 300 N = 0

T₁ = 100 N