Two packing crates of masses 10.0kg and 5.00kg

Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley as shown in the figure. The 5.00-kg crate lies on a smooth incline of angle 40.0°. Find (a) the acceleration of the 5.00-kg crate and (b) the tension in the string.

[@yvangabrielredondo9972]

somehow this was specifically the question I was looking for

[@bathong188]

Life saving video! Thank you!

[@elivertap9891]

Making sense of how to set up the equation is the most difficult part about Physics. The video helps. Thanks!

[@tamanaradmanish2249]

Thank you!!!!

[@smiley7083]

Thank you! I tried the example in my book four times. 😬

[@gl0wTech]

thank you so much. ilove you .

[@deksterr]

"how WNY Tutor saved finals week"

[@kingsleyseow4715]

how about the forces in the Y direction of the 5.00kg crate? is it negligible?

[@leas5664]

i don't see anymore recent replys. Im also wondering about the Y forces on the 5 kg crate. do they cancel each other out?

[@gabrielfriederichs7126]

ye

[@suad8289]

there was a simple law you used in one of your videos where a= forces in one box / sum of the masses is it correct? i cant find it

[@bellewaterson3474]

Why did you not use the Sumof Fx = Fn - m2g*cos40 ?

[@blessingsndovorwi1653]

Dope

[@aureeel]

Do we ignore the Y component happening in m_2 ? I noticed we didn't do anything for F_N

[@katelynjeffries634]

Why do you subtract the two equations at 7:10 instead of adding them together?

[@guggystyle]

Because if you add them, the "T" will not cancel, and you'll be stuck with 2 unknowns (T and a) instead of one unknown (a). Does that make sense?

[@tanvilodhia6718]

how do you know where to locate the angle of 40 on your free body diagram?

[@guggystyle]

Good question. It's tough to show that in a comment, but basically any time you draw a FBD of an object on an incline, the angle between mg and the direction perpendicular to the incline will ALWAYS be the same angle as that of the incline.

[@danisha9293]

This is the part that confuses me in FBD :/

[@sahilkhan-zk4em]

Suppose If in question given only find the tension in the string can we solve by applying only equilibrium condition???

[@aeronjamesmagdaloyo297]

Salamat kaayu sir, mwah mwah chupchup

[@mk-19memelauncher65]

wouldnt it be negative accel
cus its going down

[@guggystyle]

The 5.00 kg crate is accelerating UP the ramp, which we defined to be the positive direction.

[@thtchick1]

Why isn't the T negative instead of positive on the X axis, and why isn't the m2GSin40 positive?

[@guggystyle]

You are free to assign directions as either positive or negative, as long as you are consistent in the entire problem. We assigned UP the ramp as the positive direction, and since T is pointing up the ramp it is positive; the m2gsin40 is down the ramp, so it is negative.

[@ibphysicsclassroom]

where there is motion we choose that direction as positive. in this case the block will move up

[@GuadalupeBearz]

Why is the unit for tension N and not kg ? :)

[@Billy-lj3nl]

A Newton (N) is the SI unit for force, while a kilogram (kg) is a unit of mass. Since we were calculating the tension force in the string, the unit used is a Newton. A Newton is defined as kg*m/s^2, which you can see from the equation F=ma from Newton's second law. If you use kg to measure mass and m/s^2 to measure acceleration, then force = kg * m/s^2, which we call a Newton.

[@user-tt9pj5je4g]

🍪💕Thahk

[@aokosakano3037]

I thought that the components in the x direction are determined by left and right, not up and down. Why wouldn't it be left-negative and right-positive?

[@guggystyle]

The x/y axis is "tilted", so the "left" direction is also "up" the ramp.

[@ibphysicsclassroom]

where there is motion we choose that direction as positive.