Question

For a system shown in figure, force F is such that 1 kg block remains stationary relative to the wedge. There is no friction anywhere then,

• A. Force F = 7.5 N
• B. Normal by wedge on block is 6N
• C. Acceleration of the block is $\frac{15}{8}m/s^2$
• D. Tension in the string is 6.5 N

Answer 1

Answer: (A), (C)

Acceleration of the block is $\frac{15}{8}m/s^2$

let the acceleration of the system be $a$, then
$F=4a$ -- (1)

F.B.D of 1 kg w.r.t. 3 kg

By balancing the forces in $x$ direction,
$F=mgsin\theta +macos\theta$ ...(2)
By balancing the forces in $y$ direction,
$mgcos\theta =N+masin\theta$ ....(3)
Solving (1) (2) & (3)
We get T = F = 7.5 N
$a=\frac{15}{8}m/s^2$
$N=6.875$