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question

I got this question and so far I think I have worked out the tension correctly.

2T = 5g + 4g

T= 4.5g

However I don't know how to proceed with the second part where it is asking the force exerted on B by ground. I think this would be the reaction force they are asking for? Any help would be appreciated

user122343
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    Hint: draw a free body diagram for $B$. – John Douma Dec 22 '18 at 22:17
  • Also, if $T$ is $4.5$ and $P$ is $4$, how can the system be in equilibrium? – John Douma Dec 22 '18 at 22:21
  • You are adding all the forces on the system as a whole (and you also forgot to include the normal force). Consider the forces acting on $B$ and $P$ individually. – Winther Dec 22 '18 at 22:24
  • @Winther okay I get what you're saying now . I still have some questions though. Would the tension not be the same in both sides of string or is it because B is on the ground the tension won't be the same ? – user122343 Dec 22 '18 at 22:27
  • No the tension is the same at both places (the pulley is smooth so there is no friction that could change that). – Winther Dec 22 '18 at 22:28
  • Ahh. Okay so therefore the tension is T= 4g and the reaction,R =1g ? – user122343 Dec 22 '18 at 22:30
  • Yes. Also when doing problems like this it's also useful to consider special cases where you know what happens to check your answer. If you make an expression for the normal force as function of the two masses (call them $m_B$ and $m_P$) then for example if $B$ and $P$ weigh the same then there should be no normal force (they can balance) and if $P$ is heavier than $B$ then you should get a negative force: the ground has to pull not push to keep equilibrium. – Winther Dec 22 '18 at 22:33
  • Thanks allot for your help ! – user122343 Dec 22 '18 at 22:37

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For P you would be having T=4g and for B you would have N+T=5g. Solving For N and T you would find N=1g and T=4g(trivially from first eqn.). Here N=Normal force by ground.

Manraj
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