A body of mass 10kg is placed on a smooth inclined plane as shown in the figure. The inclined plane is moved with a horizontal acceleration a.
The normal reaction between block and inclined plane is
A
92N
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B
44N
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C
56N
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D
Can't be determined
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Solution
The correct option is C56N FBD of block w.r.t. inclined plane ∑Fx=0⇒Tcos37∘−Nsin37∘−ma=0 ⇒4T−3N=200 ... (i) ∑Fy=0⇒Ncos37∘+Tsin37∘−10g=0 ⇒4N+3T=500 ... (ii)
By solving equation (i) & (ii), N=56 N