A) 20 N downwards
B) 25 N upwards
C) 15 N downwards
D) 20 N upwards
Correct Answer: A
Solution :
As shown in the figure. Normal reaction \[N=F\cos {{30}^{o}}=\frac{100\sqrt{3}}{2}\text{newton}\] \[{{f}_{\max }}=\mu N=\frac{1}{4}\times \frac{100\sqrt{3}}{2}=\frac{25}{2}\sqrt{3}\,N\] Now, net force acting on the block excluding friction \[=F\sin {{30}^{o}}-3g\] \[=100\times \frac{1}{2}-3\times 10=20\,N,\]upwords Hence, the correction option is [a].You need to login to perform this action.
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