A tow truck pulls on a car that is stuck in the mud, with a force of 2960 N as s

Question

A tow truck pulls on a car that is stuck in the mud, with a force of 2960 N as shown in Fig. P5.27. The tow cable is under tension and therefore pulls downward and to the left on the pin at its upper end. The light pin is held in equilibrium by forces exerted by the two bars A and B. Each bar is a strut; that is, each bar whose weight is small compared to the forces it exerts and which exerts forces only through hinge pins at its ends. Each strut exerts a force directed parallel to its length. Determine the force of tension or compression in each strut. Proceed as follows. Make a guess as to which way (pushing or pulling) each force acts on the top pin. Draw a free-body diagram of the pin. Use the condition for equilibrium of the pin to translate the free-body diagram into equations. From the equations calculate the forces exerted by struts A and B. If you obtain a positive answer, you correctly guessed the direction of the force. A negative answer means the direction should be reversed, but the absolute value correctly gives the magnitude of the force. If a strut pulls on a pin, it is in tension. If it pushes, the strut is in compression. Identify whether each strut is in tension or in compression. Force exerted by strut A N Force exerted by strut B N?

Explanation / Answer

Now the tension on the tow cable is 2920 N
Hence
We can write this as
Tc= - 2960sin60i - 2960cos60j

let the force on the bar a be Taj

The force in the bar b be
bcos50i +bsin50j

Hence balancing the equation for horizontal components :

- 2960sin60 + bcos50 = 0
b = 3992.77 N

Now balancing for vertical we get
- 2960cos60 + a + 3992.77sin50 = 0
a = -1473.82 N(approx)

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