Since the 7 tones load is pulled along horizontal the normal force= weight of the Load which is equal to 68600N and the acting at the of 30° by the free body diagram the horizontal force = 39606.23N this is the force which must be overcome to accelerate the load of 7000kg.
Please please please leave your units in. You don’t explain WHY the coefficient of friction doesn’t have units. It’s because the units divide out. Leaving units out is a really bad habit for students
I have my grade 11 physics dynamics test next period in a couple of minutes. I hope I do well. I missed 2 days of school because I moved homes and now I fell behind in friction and gravitation.
What if i have to calculate the coefficient of friction in an angle and with the downwards force. For example when i have a box on a sin=15° degree angle slope and the downwards force "gravity" is 200N. I have been trying to solve this for two days and i'm supposed to get the answer Gx = 51,763N. I just dont get it. The equation goes sin=15°, G=200N, Gx=Fu. Sin= Gx:G, Gx= 200N • sin15°, Gx= 51,763N. What am i missing?
a load of 7tonnes is pulled along a horizontal track by a force inclined at 30 degrees to and above the track .Find the frictional force if the coefficient of friction is 0,2..................help..??
You'll have to find the normal force (mgcostheta) then plug that into the friction force equation (Ff = mu x Fn). Looks like you'll have to convert 7 tons to kg.
I have done that 7000kg *9.8 to find the weight of the Load. And the weight of the Load= to the normal force since it is pulled along the horizontal surface.
This is gonna help me sooooo much.
Have a physics test tomorrow,
Will update.
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good job!
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Now I have a physics text tmr, will update 😳
@@charlielesch6323 it's been a month
Thx a lot for this. Helped w/ my IB physics lab report
IB physics gang!
Hi , Thankyou so much .
Lots of love 💕💕 from INDIA.
The 10 sec version Coef of Friction = Force/ wt.
For ex, if 1 lb moves a 2 lb object Coef = .5 (it takes half the wt to move it)
you saved my life, THANK YOU SO MUCH!
It's amazing explanation..!! Love from India dude ❤💗
Thanks so much. My teacher just threw me in the deep end, and this really saved me😊
Thank you ! 😃 you made it look quite easy, I am not even good at math and I understood that 😅.
Thanks 😊🙏
Thank you so much, you made it so easy for me!
thank you this is the video that im looking for two days hahahaha
It helped me a lot thanku
I am glad I watched your video
I got 100 /100 💯💯
Love you ❤you saved my life
Thank you so much for this my teacher is terrible at explaining things
Since the 7 tones load is pulled along horizontal the normal force= weight of the Load which is equal to 68600N and the acting at the of 30° by the free body diagram the horizontal force = 39606.23N this is the force which must be overcome to accelerate the load of 7000kg.
physics final thank you for the help
Saved my life
Thank youuuuu ❤️❤️❤️ I was really having troubles with this one 😭😭
Please please please leave your units in. You don’t explain WHY the coefficient of friction doesn’t have units. It’s because the units divide out. Leaving units out is a really bad habit for students
Dont put units in for coefficient of friction you lose marks
I have my grade 11 physics dynamics test next period in a couple of minutes. I hope I do well. I missed 2 days of school because I moved homes and now I fell behind in friction and gravitation.
You are very op brother
It can be obtain by inequality
شكراً لك كان مفيد جداً لي
مو عربي
Thank You!!!!!!!
Thanks for this video!
Brilliant
yes. this video was very helpful thanks
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How do you calculate the coefficient if there where two blocks with different masses connected
i love you
This video saved my liiiife oh my gaaawd😂😂
And how is that dude
@@sohanreddyk2537 maybe he had an exam and watched this vid??????
What if i have to calculate the coefficient of friction in an angle and with the downwards force. For example when i have a box on a sin=15° degree angle slope and the downwards force "gravity" is 200N. I have been trying to solve this for two days and i'm supposed to get the answer Gx = 51,763N. I just dont get it. The equation goes sin=15°, G=200N, Gx=Fu. Sin= Gx:G, Gx= 200N • sin15°, Gx= 51,763N. What am i missing?
thanks bruh
THANK YOU!
Tnx a lot sir!
thank you brother
Thnx
Thank youu❤️💯👍
Hi is it possible if you explain superposition theorem 😊
Thank you👍
thanks so much
THANK YOU
Thank you
a load of 7tonnes is pulled along a horizontal track by a force inclined at 30 degrees to and above the track .Find the frictional force if the coefficient of friction is 0,2..................help..??
You'll have to find the normal force (mgcostheta) then plug that into the friction force equation (Ff = mu x Fn). Looks like you'll have to convert 7 tons to kg.
I have done that 7000kg *9.8 to find the weight of the Load. And the weight of the Load= to the normal force since it is pulled along the horizontal surface.
Isnt that the limiting friction?
0:46
king
انا عندي امتحان و الساعة 4 الفجر و للحين ما درست😂😂
😂😂😂نفس بعض
the greek letter is called "me" :')
what if there’s a frictional force present in the box??
Friction is a force that is between 2 surfaces. so friction is between the box and the ground. There is no friction 'present in the box'
Eat your cereal
L
Thanks!!!
Thanks ☺️