Asked by God'sp
An object of mass 10kg is placed on an inclined plane at 30 degree to the horizontal. Calculate the reaction between two surface and what is the coefficient of static friction.
Answers
Answered by
Damon
normal force = m g cos 30
so
friction force </= mu m g cos 30
can only say less than if object is not moving.
component of weight down slope = m g sin 30
so if no other forces are acting
m g sin 30 </= mu m g cos 30
or
mu >/= tan 30
so
friction force </= mu m g cos 30
can only say less than if object is not moving.
component of weight down slope = m g sin 30
so if no other forces are acting
m g sin 30 </= mu m g cos 30
or
mu >/= tan 30
Answered by
Henry
M*g = 10 * 9.8 = 98 N.
Fp = 98*sin30 = 49 N.
Fn = 98*Cos30 = 84.9 N.
Fp-u*Fn = M*a.
49-84.9u = 10*0 = 0
u = 0.58
Fp = 98*sin30 = 49 N.
Fn = 98*Cos30 = 84.9 N.
Fp-u*Fn = M*a.
49-84.9u = 10*0 = 0
u = 0.58
Answered by
jessica
please throw more light in this question please.
Answered by
Isioma
Reaction btwn surfaces :R=W cos รจ
Was not fully explained.
Was not fully explained.
Answered by
Wealth
Dis doesn't seem like an answer
Answered by
ja
An object of mass 10kg is placed on inclined surface 30 degree to horizontal surface .Find Normal force ?
Normal force
N
Normal force
N
Answered by
ja
An object of mass 10kg is placed on inclined surface 30 degree to horizontal surface .Find Normal force ?
Normal force
N
Normal force
N
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