Answer:
1250 N
Explanation:
length of wrench = 20 cm = 0.2 m
Write the length of wrench in vector form
r = 0.2 j metre
Direction of force= [tex]\frac{0i+4j-3k}{\sqrt{0 + 16 + 9}}=\frac{0i+4j-3k}{5}[/tex]
Let the magnitude of force is F
So, write the force in vector form
F = F [tex]\frac{0i+4j-3k}{\sqrt{0 + 16 + 9}}=\frac{0i+4j-3k}{5}[/tex]
torque, T = 150 Nm
Torque = T = r x F
[tex]150 i =0.2j\times F\frac{0i+4j-3k}{5}[/tex]
F = 1250 N
Thus, the magnitude of force is 1250 N.