PLUS Halliday, Fundamentals of Physics, 10e Calculus-based Physics I & II (PH 20

Question

PLUS Halliday, Fundamentals of Physics, 10e Calculus-based Physics I & II (PH 201-202 tudy & Assignment ORION t Open Assignment SOURCES a) Number (b) Number c) Number d) Number by Study FULL SCREEN PRINTER VERSION .BACK NEXT Chapter 12, Problem 010 The system in the figure below is in equilibrium, with the string in the center exactly horizontal. Block A weighs 45.0 N, block B weighs 56.0 N, and angle p is 39.0 What are Find (a) tension T1, (b) tension T2 (c) tension T3, and (d) angle 6. units Units units click if you would like to show Work for this question

Explanation / Answer

the center of the figure is horizontal means it is providing no vertical force. All the tension that holds up each block comes from the angled strings at the left and right.

For the blocks to be stationary, all the forces on them must sum to 0, so the upward force supplied by the string is just the same as the weight force of the block, which we are given. We need to find what the total tension in string 1 needs to be in order to have a vertical component of 45 N. T1 is the hypotenuse of a right-angled triangle, and the opposite side to the angle (39°)is the horizontal force, the adjacent side is the vertical force. We know from trig that:

cos=adjacent/hypotenuse

Rearranging and substituting in:

T1=vertical force/cos=45/cos39=57. 9 N

Using the same kind of reasoning, the horizontal force on this string, which is equal to T2 will be:

T2=T1sin=57. 9sin39=36.44 N

This will also be the horizontal component of the force provided by string 3 on the second mass:

sin=T2/T3 : .............................call this Equation 1

We know T2 but don't yet know either or T3. We can use the vertical force on block B to find both. We do know that the weight of block B is 56 N.

T3=vertical force/cos

T3=56/cos

Substitute this value for T3 into Equation 1:

T2/(56/cos) = sin

Invert and multiply:

T2 x (cos/56) = sin

Substitute in the known value of T2 and rearrange:

sin/cos = 36.44/56

tan = 36.44/56

=tan-1(36.44/56)=33.05°

We can then use this in Equation 1 to find the value of T3:

sin=T2/T3

Rearranging and substituting in the value of T2:

T3 = 36.44/sin33.05°= 66.81N

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