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21. A lift of mass Mis suspended from a cable. The lift descends with a downward acceleration, a. A frictional force Facis on the lift. What is the tension Tin the cable? A T=Ma+F B T=Ma-F C T=M(g+a)-F D T=M(g-a)-F

Question

21.
A lift of mass Mis suspended from a cable. The lift descends with a downward acceleration, a. A
frictional force Facis on the lift.
What is the tension Tin the cable?
A T=Ma+F
B T=Ma-F
C T=M(g+a)-F
D T=M(g-a)-F

21. A lift of mass Mis suspended from a cable. The lift descends with a downward acceleration, a. A frictional force Facis on the lift. What is the tension Tin the cable? A T=Ma+F B T=Ma-F C T=M(g+a)-F D T=M(g-a)-F

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PryderiProfessional · Tutor for 6 years

Answer

D \( T=M(g-a)-F \)

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## Step1:<br />First, we need to understand the forces acting on the lift. The lift is moving downwards, so the forces acting on it are its weight, which is \( M \times g \) (where \( g \) is the acceleration due to gravity), the tension \( T \) in the cable pulling it upwards, and the frictional force \( F \) acting upwards.<br /><br />## Step2:<br />According to Newton's second law, the net force acting on an object is equal to its mass times its acceleration. In this case, the net force acting on the lift is the difference between the weight of the lift and the sum of the tension in the cable and the frictional force. This can be written as:<br /><br />### \( M \times g - (T + F) = M \times a \)<br /><br />## Step3:<br />Rearranging the equation to solve for \( T \), we get:<br /><br />### \( T = M \times g - M \times a - F \)<br /><br />This equation represents the tension in the cable.
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