- #1
- 374
- 0
Force Body Diagram Hamster-Please Help!
A 200 g hamster sits on an 800 g wedge shaped block. The block, in turn rests on a spring scale.
a) Initially static friction is sufficient to keep the hamster from moving. In this case the hamster and the block are effectively a single 1000g mass and the scale should read 9.8 N. Show that this is the case by treating the hamster and the block as separate objects and analyzing the forces.
I am reviewing, and I know I did this problem correctly before, but now it's not working for me. Could someone please tell me what I am doing wrong
Hamster: I drew a force body diagram, tilting the x & y axes
Sum of forces on y = Force of Block on hamster - mgcos40=0
Force of block on hamster = (.200kg)(9.80)(cos40)
Block: For the block I didn't tilt the coordinates
Sum of forces on y= normal force - Force of hamster on block in y direction - Mg
= n - (.200kg)(9.80)(cos40)(cos40) - (.800kg)(9.80)
n= -8.99 N which is the answer for the second part of the question, when the static friction goes away.
Homework Statement
A 200 g hamster sits on an 800 g wedge shaped block. The block, in turn rests on a spring scale.
a) Initially static friction is sufficient to keep the hamster from moving. In this case the hamster and the block are effectively a single 1000g mass and the scale should read 9.8 N. Show that this is the case by treating the hamster and the block as separate objects and analyzing the forces.
Homework Equations
The Attempt at a Solution
I am reviewing, and I know I did this problem correctly before, but now it's not working for me. Could someone please tell me what I am doing wrong
Hamster: I drew a force body diagram, tilting the x & y axes
Sum of forces on y = Force of Block on hamster - mgcos40=0
Force of block on hamster = (.200kg)(9.80)(cos40)
Block: For the block I didn't tilt the coordinates
Sum of forces on y= normal force - Force of hamster on block in y direction - Mg
= n - (.200kg)(9.80)(cos40)(cos40) - (.800kg)(9.80)
n= -8.99 N which is the answer for the second part of the question, when the static friction goes away.