A meter stick on a horizontal frictionless table top is pivoted at the 80-cm mar

Question

A meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark. A force is applied perpendicularly to the end of the stick at 0 cm, as shown. A second force (not shown) is applied perpendicularly at the 100-cm end of the stick. The forces are horizontal. If the stick does not move, the force exerted by the pivot on the stick:


A. must be zero
B. must be in the same direction as and have magnitude
C. must be directed opposite to and have magnitude
D. must be in the same direction as and have magnitude
E. must be directed opposite to and have magnitude

Explanation / Answer

so you have a net at the pivot point is equal to zero and so is the Fnet.

Let say Fo is force at do=0 cm and F1 is the force at d1=100cm. Fp is the force at the pivot point. So you have. Let say force up is positive and force down is positive. Since the system is in equilibrium you have:

Fnet= Fp-(Fo+F1)=0 ---> Fp=F0+F1 so the force at the pivot point has to be directly opposite and has the same magnitude of the sum of Fo and F1.! Goodluck!

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