Respuesta :
The mass M1 is 7.8 kg
Explanation:
Block M1 is hanging on the string while block M2 is on the frictionless ramp.
We have to write the equations of motion for the two blocks.
- For M1, the only two forces acting on it are the force of gravity [tex]M_1 g[/tex] (downward) and the tension in the string T (upward). So we can write
[tex]M_1 g - T = M_1 a[/tex]
where
[tex]M_1[/tex] is the mass of the block
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]a[/tex] is the acceleration of the system
- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, [tex]M_2 g sin \theta[/tex]. So the equation of motion is
[tex]T-M_2 g sin \theta = M_2 a[/tex]
where
[tex]M_2 = 13.5 kg[/tex] is the mass of the 2nd block
[tex]\theta=35.5^{\circ}[/tex] is the angle of the ramp
In order for the two blocks to be in equilibrium, the acceleration must be zero:
[tex]a=0[/tex]
So the two equations become:
[tex]M_1 g - T=0\\T-M_2 g sin \theta = 0[/tex]
Isolating T from the 1st equation,
[tex]T=M_1 g[/tex]
And substituting into the 2nd equation, we can find the value of the mass [tex]M_1[/tex]:
[tex]M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg[/tex]
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The value of mass of block M1 for which the two blocks are in equilibrium is 7.84 N.
What is tension in the string due to hanging weight?
Tension is the pulling force carried by the flexible mediums like ropes, cables and string. Tension in a body due to the weight of the hanging body is the net force acting on the body.
[tex]T=ma[/tex]
Here, (m) is the mass and (a) is the acceleration.
The tension in the string due to gravitational force and acceleration force by mass M1 is,
[tex]T=M_1(g+a)[/tex] ......1
As, mass M2 is, rests on a long ramp of angle 35.5°. Thus, the tension in the string due to gravitational force and acceleration force by mass M12 is
[tex]T=M_2(gsin(35.5)+a)\\T=13.5(9.81\sin(35)+a)[/tex]
Now the two block are in equilibrium (not accelerating) position. Thus, the acceleration should be zero. Therefore,
[tex]T=13.5(9.81\sin(35.5)+0)\\T\cong76.905\rm N[/tex]
Put this value of tension in the equation 1 as,
[tex]76.905=M_1(9.81+0)\\M_1=\dfrac{200}{9.81}\\M_1=7.84\rm N[/tex]
Hence, the value of mass of block M1 for which the two blocks are in equilibrium is 7.84 N.
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