- #1

fishiiie

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## Homework Statement

A block of mass m = 0.570 kg rests on top of a block of mass M = 2.30 kg. A string

attached to the block of mass M is pulled so that its tension is T = 6.20 N at a 20 degree angle to

the horizontal. The blocks move together. The coefficient of static friction at

the surface between the blocks is μs = 0.43; there is no friction at the surface between

block M and the floor.

a) What is the direction of the frictional force being exerted on the top block of mass m?

Justify your answer.

b) The tension T is now increased - what is the maximum tension, T

_{max}, with which the string can be pulled such that the blocks continue to move together (i.e. that the block of

mass m does not start to slide on top of the block of mass M)?

## Homework Equations

F

_{net}= ma

F

_{smax}≤ μF

_{n}

## The Attempt at a Solution

a) (Just getting some clarifications on this one to see if I did it right.)

I wrote that friction force should be directed to the left, since static friction (and kinetic friction) always opposes the direction of motion/acceleration.

b) So I'm just trying to draw FBDs for the two blocks and I'm getting confused just doing it. I can't even get the answer if I can't get the FBDs right, so please help me :(

__FBDs__:

Top block (m):

ƩF

_{x}: -Fs = ma

ƩF

_{y}: Fn = mg

Bottom block (M):

ƩF

_{x}: Tcos20 - Fs = Ma

ƩF

_{y}: Fn + Tsin20 = Mg

I'm assuming that both blocks have a normal force on them, but I'm not so sure because I'm getting very confused between normal force and contact force... (Newton's Third Law)

And I'm not sure about the frictional force between the two blocks in the FBDs.. Since it's acting on both blocks (because the friction is between the two blocks), it is included in both of the blocks' FBDs.. right?