Wall by Force F - F -60 N- find friction force between block - wall- - s -0-8- - k -0-7A- 48 NB- 42 NC- 40 ND- 36 N

For equilibrium along direction ⊥ to wall F=N N=60 f_max=μ_sN=0.8 × 60=48 Force || to surface ⇒ mg=40 N which is less than limiting friction. Hence, f=40 ...

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Standard XII

Physics

Normal Reaction

wall by Force...

Question

Block A is pressed against the wall by Force F, F=60 N, find friction force between block & wall. ( $μ_{s} = 0.8, μ_{k} = 0.7$ )

48 N

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42 N

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40 N

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36 N

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Solution

The correct option is C 40 N

For equilibrium along direction $⊥$ to wall

F=N

N=60

$f_{m a x} = μ_{s} N = 0.8 \times 60 = 48$

Force || to surface $\Rightarrow m g = 40$ N which is less than limiting friction. Hence,

f=40 N

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Block A is pressed against the wall by Force F, F=60 N, find friction force between block & wall. (μs=0.8,μk=0.7)

The correct option is C 40 N For equilibrium along direction ⊥ to wall F=N N=60 fmax=μsN=0.8×60=48 Force || to surface ⇒mg=40 N which is less than limiting friction. Hence, f=40 N

Block A is pressed against the wall by Force F, F=60 N, find friction force between block & wall. ( $μ_{s} = 0.8, μ_{k} = 0.7$ )

The correct option is C 40 N

For equilibrium along direction $⊥$ to wall

F=N

N=60

$f_{m a x} = μ_{s} N = 0.8 \times 60 = 48$

Force || to surface $\Rightarrow m g = 40$ N which is less than limiting friction. Hence,

f=40 N