07-Newton\'s Laws: Tension and Pulleys Begin Date: 9/25/2017 8:00:00 AM Due Date

Question

07-Newton's Laws: Tension and Pulleys Begin Date: 9/25/2017 8:00:00 AM Due Date: 9/29/2017 7:50:00 AM End Date: 10/3/2017 7:50:00 A (8%) Problem 1: A man is attempting to lift a crate using a two part pulley systen as shown in the image. The crate has mass my m 92 kg and the man has m 75 kg. He pulls downward on the rope with a force of magnitude F- 608 N. The pulleys are massless and frictionless. m2 Otheexpertta.com 25% Part (a) Using T to describe the ma on the crate in terms of gravity and the variables provided. gnitude of the tension force, write an expression for the sum of the forces in the y-direction acting -| Grade Summary Deductions 92% Submissions Attempts remaining (7%, per attempt) etailed view 112 Subenit 1 give up! Hints: 3 for a deduction Hints remaining Feedback:--deduction pet feedback. Draw a Free Body Diagram of the crate. How many forces act on the crate? s the crate's weight balanced by one or two portions of rope? Is the tension the same in both of the portions of the rope holding the crate? 25% Part (b) Using the results from above, write an expression for the crate's vertical acceleration, ao in terms of T 25% Part (c) what is the magnitude of the tension force, T in newtons? 25% Part (d) what is the block's acceleration, de, in me? for the crate's vertical acceleration, a, in terms of T

Explanation / Answer

A)

Forces on the crate are:

i) two tension forces(T) in the string acting upwards

ii)weight (m2*g) acting downwards

So, sum of all forces in y-direction:

2T - m2*g

B)

Now, 2T - m2*g = m2*ax

so, ax = (2T - m2*g)/m = 2T/m2 - g

C)

Tension force = pull of the rope = 608 N

D)

ax = 2T/m2 - g

= 2*608/92 - 9.8

= 3.42 m/s2

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