Question

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

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 \(g=9.8\space m/s^{2}\). 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 (net force \(F_{net}=0\)), the total tension \(T_{total}\) in the ropes balances the weight of the crate. If the tension in each rope is \(T\), and there are \(n = 4\) ropes, then \(T_{total}=nT\). So, \(T=\frac{W}{n}\). Substituting \(W = 196\space N\) and \(n = 4\), we get \(T=\frac{196}{4}=49\space N\).

Answer:

B. \(49\space N\)