A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by [2001]

Question

Smart Achievers · Accepted Answer

(4)

$At equilibrium T = M g$

$F.B.D. of pulley$

$F_{1} = (m + M) g$

$The resultant force on pulley$

$F = \sqrt{F_{1}^{2} + T^{2}} = [\sqrt{{(m + M)}^{2} + M^{2}}] g$

$As pulley is on rest, force applied by clamp should be equal to F and opposite to it.$

Solution

Q.
A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by [2001]

1 $\sqrt{2} M g$

2 $\sqrt{2} m g$

3 $\sqrt{{(M + m)}^{2} + m^{2}} g$

4 $\sqrt{{(M + m)}^{2} + M^{2}} g$

Ans.
(4)

$At equilibrium T = M g$

$F.B.D. of pulley$

$F_{1} = (m + M) g$

$The resultant force on pulley$

$F = \sqrt{F_{1}^{2} + T^{2}} = [\sqrt{{(m + M)}^{2} + M^{2}}] g$

$As pulley is on rest, force applied by clamp should be equal to F and opposite to it.$

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Solution

Q. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by [2001]

1 2 Mg

2 2 mg

3 (M+m)2+m2g

4 (M+m)2+M2g

Ans. (4) At equilibrium T=Mg F.B.D. of pulley F1=(m+M)g The resultant force on pulley F=F12+T2=[(m+M)2+M2]g As pulley is on rest, force applied by clamp should be equal to F and opposite to it.