Solutions must be worked out symbolically first (that is, in terms of variables

Question

Solutions must be worked out symbolically first (that is, in terms of variables and constants) with numbers plugged in as the very last step. Make sure to double check the answer!

You are trying to pull open a door that is stuck shut. To do this, you apply a force to the doorknob in a direction perpendicular to the door, but you are not successful. Your friend suggests that you can exert more torque by tying a rope to the doorknob and pulling with the same force as before. Is your friend correct?

You have a long rod that is allowed to pivot around its center of mass. You hang a weight with mass m1 at a distance d to the left of the pivot point. When you place a mass m2 a distance of 2d to the right of the pivot point, the rod remains horizontal. What is the relationship between the two masses?

You hang a 150 g meterstick from a string tied around it at the 40cm mark. From which mark on the meterstick do you need to hang a 50g mass for the meterstick to remain horizontal? Assume the meterstick has a uniform mass distribution.

Explanation / Answer

Answer 1

No ,He is not correct as torque=force*perpendicular distance

Since tension in the rope =force applied

also perpendicular distance will also remain same

so torque will remain same .So friend is not correct

Question 2

writing torque about center of mass

m 1*d-m 2*2 d=0

we got

m 1=2 m 2

(3)

given mass of meter stick=Mm=150 g

Let the mass is hung at the distance of x meter from the center of meter stick

Since meter stick is horizontal,net torque must be zero around any point

calculating torque about center of meter stick

Let tension in the string be T N and distance of string from midpoint of meter stick=50-40 =10 cm

50*x-T*10=0

we got

x=T*10/50

Also balancing force

T=150+50 =200

we got x=200*10/50=40 cm from midpoint so

at 90 cm mark,we must hang 50 g object

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