2. Consider the following system with a 2 kg mass sitting on a table, connected

Question

2. Consider the following system with a 2 kg mass sitting on a table, connected to both 1 kg and 3 kg masses by strings which go over two separate pulleys and are pulling on the 2 kg mass in opposite directions.

What static friction coefficient would be required between the 2kg mass and the table to stop the masses from accelerating?

If the masses are in motion, what kinetic friction coefficientwould be required between the 2 kg mass and the table to ensure the masses move at a constant velocity?

Find the tension in both pieces of string for the case of zero friction.

Explanation / Answer

part 1:

when the masses are not moving, the tension in the strings will be equal to weight of the blocks attached to the end of the strings.

lets assume that 1 kg block is connected on the left hand side and 3 kg block is connected on the right hand side.

then left hand side string tension=1*9.8=9.8 N

right hand side string tension=3*9.8=29.4 N

so the friction force will act along the tension in left side and oppose the tension in right side.

let static friction coefficient be k.

then 9.8+k*2*9.8=29.4

k=1

hence static friction coefficeint has to be 1 in order to ensure that the block does not move.

part 2:

for the blocks to move in uniform velocity, acceleration=0

net force on the 2 kg block is 0.

let the kinetic friction coefficient=k

then 29.4-9.8-k*2*9.8=0

k=1

tensions willbe 9.8 N and 29.4 N respectively for string connected to 1 kg and 3 kg blocks .

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.