Posted by billy on Wednesday, February 16, 2011 at 12:04am.
Two blocks are arranged at the ends of amassless string as shown in the figure. The system
starts from rest. When the 1.67 kg mass has fallen through 0.395 m, its downward speed is
The acceleration of gravity is 9.8 m/s2.
What is the frictional force between the
3.34 kg mass and the table?
college physics - drwls, Wednesday, February 16, 2011 at 5:21am
I don't understand the figure, since it isn't shown. Is 3.34 kg on a flat horizontal surface and is 1.67 kg pulling it over to the edge? Does the string go over a pulley? With no pulley, one would expect a lot of friction where the string goes over the edge.
Both masses accelerate at a rate a given by
V^2 = 2 a X
a = V^2/(2X) = 1.658 m/s^2
Write free body diagram equations for both masses, eliminate the spring tension and solve for the friction force
college physics - billy, Wednesday, February 16, 2011 at 6:07am
the 3.34 is on a horizontal table and the 1.67 is over the edge. they are attached by a string and I need to know the friction of the horizontal mass (3.34 kg) has on the table with the vertical mass (1.67 kg) hanging off the table.
college physics - drwls, Wednesday, February 16, 2011 at 6:27am
To solve this you have to know something about the additional friction where the string goes over the edge. Perhaps they want you to assume it is zero. That is why I asked if there is a pulley wheel at the edge.
If there is no string friction,
1.67 g - T = 1.67 a
T - Ff = 3.34 a
1.67 g - Ff = 5.01 a
I already solved for the acceleration a.
Ff = 1.67 g - 5.01 a = 8.06 newtons
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