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# Normal Reaction Force Mechanics Problem :'( Watch

1. Hi, in advance thanks for the help.

I am stuck on a mechanics question:

Parcel of mass 5kg rests in a plane inclined 45 degrees to the horizontal.
Parcel held in equilibrium by magnitude 20N inclined 30 degress to slope. Show hat normal contact force between parcel and plane is 24.6N!

Now i see that mgsin(45) = 34.64 and 20sin(30) = 10, but why do you find the difference for the normal reaction force? The answer is there, 34.6-10 but why do you minus it? I thought that 34.6N + 10 - 49 = R?

Thanks for help
2. The weight of the mass perpindicular to the slope is mg.cos(45)

Supporting the mass is the normal reaction from the slope. Also supporting the mass, is a perpindicular component from the 20N tension, which is 20.sin(30) = 10N.

So the supporing forces are NR + 10N and the weight of the mass (perpindicular to the slope) is mg.cos(45).

Balancing forces gives: Nr + 10 = mg.cos(45). i.e.

Nr = mg.cos(45) - 10
===============
3. (Original post by SharpSchool)
Hi, in advance thanks for the help.

I am stuck on a mechanics question:

Parcel of mass 5kg rests in a plane inclined 45 degrees to the horizontal.
Parcel held in equilibrium by magnitude 20N inclined 30 degress to slope. Show hat normal contact force between parcel and plane is 24.6N!

Now i see that mgsin(45) = 34.64 and 20sin(30) = 10, but why do you find the difference for the normal reaction force? The answer is there, 34.6-10 but why do you minus it? I thought that 34.6N + 10 - 49 = R?

Thanks for help
The 10N is opposite to the 34.64N and considering this positive
then the force of 10N is negative.
It is because the direction of force holding the parcel is upward
inclined 30 degrees to slope, so its normal is opposite to that of
gravity force.
4. resolve parallel and perpendicular to the plane downwards since the object will slide down unless held in equilibrium

R perpendicularly down the slope gives R + 20sin30 = 5 gsin45

which gives R = 24.6

always draw a diagram and show an arrow with the direction of motion (or assumed motion if its in equilibrium) ( then resolve in that direction - parallel and perpendicular to the slope ) if i had time i'd have drawn it out and double checked but i''ve just tried picturing it in my head , forgive me if im wrong
5. Thanks a lot for your help guys . I have a test on this today so this will help!

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