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The 50 kg h...

Question

The $50 k g$ homogeneous smooth sphere rests on the $30^{o}$ incline $A$ and bears against the smooth vertical wall $B$ . The contact force at $B$ is $(g = 10 {m / s}^{2})$

250 N

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Z e r o

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\frac{500}{\sqrt{3}} N

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500 N

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Solution

The correct option is C $\frac{500}{\sqrt{3}} N$
$\sum F_{y} = 0 W = R_{A} c o s 30 R_{A} = \frac{W}{c o s 30} \sum F_{x} = 0 R_{A} = R_{A} s i n 30 = \frac{W}{c o s 30} s i n 30 = \frac{500}{\sqrt{3}} N$

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Similar questions

Q. A

50 k g

homogeneous smooth sphere rests on the

30^{\circ}

inclined plane

A

and bears against the smooth vertical wall

B

. The contact force at point

B

is
(Take

g = 10 m / s^{2}

)

Q. The

50 k g

homogeneous smooth sphere rest on the

30^{\circ}

incline

A

and bears against the smooth vertical wall

B

. Calculate the contact forces at

A

and

B

Q. A

50 k g

homogeneous smooth sphere rests on the

30^{\circ}

inclined plane

A

and bears against the smooth vertical wall

B

. The contact force at point

B

is
(Take

g = 10 m / s^{2}

)

Q. The

50 k g

homogeneous smooth sphere rests on the incline

A

and bears the smooth vertical wall

B

. calculate the contact forces at

A

and

B

Block $A$ of mass $4 k g$ is to be kept at rest against a smooth vertical wall by applying a force $F$ as shown in fig. The force required is $(g = 10 m / s^{2})$

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The 50 kg homogeneous smooth sphere rests on the 30o incline A and bears against the smooth vertical wall B. The contact force at B is (g=10m/s2)

The correct option is C 500√3N∑Fy=0W=RAcos30RA=Wcos30∑Fx=0RA=RAsin30=Wcos30sin30=500√3N

The $50 k g$ homogeneous smooth sphere rests on the $30^{o}$ incline $A$ and bears against the smooth vertical wall $B$ . The contact force at $B$ is $(g = 10 {m / s}^{2})$

The correct option is C $\frac{500}{\sqrt{3}} N$
$\sum F_{y} = 0 W = R_{A} c o s 30 R_{A} = \frac{W}{c o s 30} \sum F_{x} = 0 R_{A} = R_{A} s i n 30 = \frac{W}{c o s 30} s i n 30 = \frac{500}{\sqrt{3}} N$