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baddinlol
Group Title
A block of mass 10kg is pulled along a horizontal plane by a force of 10N inclined at 30* to the plane. The coefficient of friction between the block and the plane is 0.05, Find the acceleration of the block.
 2 years ago
 2 years ago
baddinlol Group Title
A block of mass 10kg is pulled along a horizontal plane by a force of 10N inclined at 30* to the plane. The coefficient of friction between the block and the plane is 0.05, Find the acceleration of the block.
 2 years ago
 2 years ago

This Question is Closed

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
dw:1327647830040:dw
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Wrong group. There is a physics group though. Post your question there.
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
It is in my mathematics books though.... sorry
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
well, F = m*a Force in the horizontal direction is 10*cos(30) = 5sqrt3 not sure where the coefficient of friction goes though...been awhile since physics
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
This question is acceptable here  don't worry. But you have more like in the physics group.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Luck, rather.
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
Do I subtract the friction coeffecient from the opposing force? I am confused...
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
I will try the physics one if it doesn't work out =)
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
I believe you need to multiply the normal force by the friction coefficient.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
And doing that will tell you the frictional force. You add that frictional force as a force vector, if I remember correctly.
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
hmm i think i got it now the vertical force = 10*sin(30) = 5 the downward force = 10g = 98 therefore R = 98  5 = 93 R* friction = 93*.05 = 4.65 but thats going in opposite direction net forward force = 5sqrt3  4.65 = 4.01 now apply F=ma 4.01 = 10a a = 0.401
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
Thanks, that looks right.
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
Can I do this in ij notation?
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
i guess, you could write the vectors in ij notation
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot, really helped
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
\[F_n=mg \cos \theta =(10kg)(9.81\frac{N}{kg})(\cos(30 \deg))=85.N\]\[F_f = μN\]\[F_f=(0.05)(85.N)=4.2N\]\[F=85.N4.2N=80.8N\]\[F=ma\]\[80.8N=(10kg)a \rightarrow a=8.08\frac{N}{kg}=8.08\frac{m}{s^2}\]This is what I get. I don't think you found the force normal correctly, crow.
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
oh oops, i thought that the Net force had to be 0 downward force, mg = 98N so upward force must be 98N, no?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
If the net force were 0, the block wouldn't be accelerating would it?
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
no i was speaking of vertical forces, the block is staying on the ground. isn't the only Force pushing it forward is the 10N applied at the 30* angle ?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Yes, and so it wouldn't make sense for the normal force to be more than 10. Is the 10N even enough to overcome gravity? \[F_n=mgcos \theta=(10kg)(9.81\frac{N}{kg})(\cos(30\deg))=85.0N\]
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
i dunno, that doesn't seem right the normal force is pointing straight up because the block is moving on a horizontal plane so you don't need to multiply mg by cos* the normal force will just be enough to cancel gravity > Fn = mg  10sin(3)
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Oh! I misread this question, and the picture only confused me more. I was thinking of it as a 30degree incline. Geez. Ok.
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.2
ahh, yeah so you're right if the block is moving up the incline
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Yep, and your old solution is correct :P
 2 years ago
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