Question

An object of mass 10 kg is about to slip on a plane inclined at an angle of 20 with the horizontal plane. How much upward force along the ground will apply to the object?​

Answer 1

Answer:

given below

Explanation:

mgsin370=10×10×3/5=60N

fl=μ=0.2mgcos370

=0.2×10×10×4/5

=16N

Friction force alone is unable to balance mgsin370. Hence, F is required for balancing. Here two cases arise.

Case I: 

If the block has tendency to slide down, friction on block will act upwards.

F + 16 = 60

RightarrowF=44N

This is the minimum force required for balancing.

Case II: If the block has on tendency to slide up, friction on block will act downwards.

F = 60 + 16⇒ F= 76 N

This is the maximum force required for balancing.

Therefore, for balancing 44N≤F≤16N

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Answer 2

Answer:

there should be 10 on top 10 on bottom

Explanation:

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