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In an arrangement shown below, the pulley is light and the string is light and inextensible, the contact force between ground and the block is F. If the system is in equilibtium and (g=10 m/s2), the value of F (in Newtons) will be

(Given F is resultant of N and R)

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Solution

Step 1: Draw FBD of 7 kg mass
Step 2: Using Newton's law of motion along x and y directions
ΣFx=max ; ΣFy=may
Step 3: For system to be in equilibrium (ax=ay=0)
ΣFx=0 ; ΣFy=0
Step 4: As per the given condition the value of T is 302 N
Hence, along x direction
R=Tcos45°
Along y direction
N+Tsin45°=7g
Step 5: N=70302×12=40 N
R=302×12=30 N
Step 6: The resultant of N and R is F
F=N2+R2=402+302=50 N
F=50 N

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