Block A has a mass of 6.90 kg and is on a rough incline of 18.0o to the horizontal. Block B has a mass of 5.00 kg and the coefficient of kinetic friction between Block A and the plane is 0.236. What is the acceleration of the blocks? What is the tension in the string?
The drawing looks like a right triangle, the right angle is on the right side so block A is being pulled up to the right. Theta is the angle opposite of the right.
The drawing looks like a right triangle, the right angle is on the right side so block A is being pulled up to the right. Theta is the angle opposite of the right.
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The frictional force is found by resolving the gravitational force on block A
it is coefficient * normal reaction
= 0.236 * 6.90*9.8 * cos( 18 deg) Call this Fr ( about 20 N)
The component of gravity along the plane is mg sin(theta)
= 6.90 * 9.8 * sin(18 deg) call this Fa
And the other weight is pulling down with a force of mg = 5.00 * 9.8 Call this Fb
Now it is clear that Fb is greater than Fa so the weight A is being pulled UP the slope.
Hence friction is acting DOWN the slope to oppose the motion.
The net unbalance force is then F = Fb -Fa - Fr
and of course F = ma where m is the total mass being accelerated.
Finally to get the tension in the string you can start with either weight. I would start with block B
You know that the net force on B is = Mb * a
and it is created by Mb *g ( down) - T ( up)
T = Mb * g - Mb * a
= Mb *(g-a) and you can substitute the value you calculated for a above
it is coefficient * normal reaction
= 0.236 * 6.90*9.8 * cos( 18 deg) Call this Fr ( about 20 N)
The component of gravity along the plane is mg sin(theta)
= 6.90 * 9.8 * sin(18 deg) call this Fa
And the other weight is pulling down with a force of mg = 5.00 * 9.8 Call this Fb
Now it is clear that Fb is greater than Fa so the weight A is being pulled UP the slope.
Hence friction is acting DOWN the slope to oppose the motion.
The net unbalance force is then F = Fb -Fa - Fr
and of course F = ma where m is the total mass being accelerated.
Finally to get the tension in the string you can start with either weight. I would start with block B
You know that the net force on B is = Mb * a
and it is created by Mb *g ( down) - T ( up)
T = Mb * g - Mb * a
= Mb *(g-a) and you can substitute the value you calculated for a above
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HI! To solve this question, you may need to use a vector diagram. Go ahead and draw it out! I may be wrong... But there seem to be some missing quantities like Resultant force..