The mass of an object with pulleys

[Francis Hannah]

A block is lifted by the pulley. Where θ=37.9 degrees and F=112N, find the mass of the block in kg.

I know that tension T1=T2=T3.

So, T3=112⋅cos(7.1)=111.14

Hence, T1 and T2 are equal to 111.14 as well.

Thus, mg=(2)⋅(111.14)⋅cos(37.9)=175.40?

And thus the mass is < 175.40/9.81 ⇒ mass < 17.88 kg?

Thanks for your help :)

 

[Orodruin]

What does the 7.1 degree angle have to do with anything?

 

[Francis Hannah]

What does the 7.1 degree angle have to do with anything?

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Wouldn't T3 have to be at the same angle as T1 and T2?

 

[Orodruin]

No. Any particular reason why you think so?

 

[Francis Hannah]

No. Any particular reason why you think so?

Oh, does the angle not determine how much force is required to lift the block?

 

[Orodruin]

Oh, does the angle not determine how much force is required to lift the block?

No. Again, why do you think it would?

Now, the angle $\theta$ will influence the force. I suggest you draw a free body diagram for the mass.

 

[Francis Hannah]

No. Again, why do you think it would?

Now, the angle $\theta$ will influence the force. I suggest you draw a free body diagram for the mass.

I see, thank you for your help.

So mg is equal to (2)(112)cos(37.9)?

 

[Orodruin]

Yes. For the future, please state your reasoning more clearly. It will make it easier to help you. It will also generally help if you answer direct questions.