Question

22 a 20 kg crate is suspended from a fixed beam by four separate vertical ropes. what is the approximate tension in each rope? a 800n b 5n c 49n d 200n e 80n

Explanation:

Step1: Calculate the weight of the crate

The weight \(W\) of an object is given by \(W = mg\), where \(m = 20\space kg\) and \(g\approx9.8\space m/s^{2}\). So, \(W=20\times9.8 = 196\space N\).

Step2: Determine the tension in each rope

Since the crate is suspended by four vertical ropes and is in equilibrium (not accelerating), the total upward force (sum of tensions \(T\) in each rope) equals the weight of the crate. Let the tension in each rope be \(T\). Then \(4T=W\). Solving for \(T\), we get \(T=\frac{W}{4}\). Substituting \(W = 196\space N\), \(T=\frac{196}{4}=49\space N\).

Answer:

C. 49 N